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Physics Quiz

Physics Quiz: Evaluate Device Energy Efficiency

Practice Evaluate Device Energy Efficiency in Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

A heat engine takes in thermal power $P_{in}=900\,\text{W}$ from burning fuel and produces useful mechanical power $P_{out}=270\,\text{W}$ . The rest is waste heat to the surroundings. What is the engine’s efficiency (as a percentage)?

What this quiz covers

This quiz focuses on Evaluate Device Energy Efficiency, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$30\%$ (correct answer)
$70\%$
$3.0\%$
$333\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this device, the input power is P_in = 900 W and the useful output power is P_out = 270 W, so efficiency η = (P_out/P_in) × 100% = (270/900) × 100% = 30%. The remaining power P_waste = 900 - 270 = 630 W is lost as thermal energy, which is characteristic of heat engines where incomplete combustion and friction occur. Choice A is correct because it properly calculates efficiency as (output/input) × 100%. Choice D is wrong because it inverts the ratio calculating input/output instead of output/input, giving >100% which violates conservation. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 2

Two light bulbs produce light (useful) and heat (waste).

Incandescent bulb: $P_{in}=100\ \text{W}$ electrical, $P_{out}=5\ \text{W}$ light.
LED bulb: $P_{in}=20\ \text{W}$ electrical, $P_{out}=8\ \text{W}$ light.

Which bulb is more efficient, and by how many percentage points (LED efficiency minus incandescent efficiency)?

Incandescent by $35$ percentage points
LED by $35$ percentage points (correct answer)
LED by $3.5$ percentage points
Incandescent by $3.5$ percentage points

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For the incandescent bulb, the input power is P_in = 100 W and the useful output power is P_out = 5 W, so efficiency η = (5/100) × 100% = 5%; for the LED, P_in = 20 W and P_out = 8 W, so η = (8/20) × 100% = 40%. The difference is 40% - 5% = 35 percentage points, with the LED being more efficient, characteristic of bulbs where incandescent loses more to heat via filament resistance while LED is more direct in light conversion. Choice B is correct because it accurately compares efficiencies showing which device converts larger fraction. Choice A incorrectly identifies the incandescent as more efficient. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 3

A heat engine takes in $E_{in}=5000\ \text{J}$ of thermal energy from fuel and produces $E_{out}=1400\ \text{J}$ of useful mechanical work. What is the efficiency of the engine (as a percentage)?

$3.6\%$
$28\%$ (correct answer)
$72\%$
$357\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this device, the input energy is E_in = 5000 J and the useful output energy is E_out = 1400 J, so efficiency η = (1400/5000) × 100% = 28%. The remaining energy E_waste = 5000 - 1400 = 3600 J is lost as thermal energy, which is characteristic of heat engines where incomplete combustion and thermodynamic limits cause losses. Choice B is correct because it properly calculates efficiency as (output/input) × 100%. Choice D inverts the ratio calculating input/output instead of output/input, giving >100% which violates conservation. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 4

A solar panel receives $E_{in}=2000\ \text{J}$ of solar radiation in a short time interval and produces $E_{out}=320\ \text{J}$ of electrical energy. The rest becomes thermal energy in the panel. How much energy is lost as waste heat, $E_{lost}$ ?

Use $E_{in}=E_{out}+E_{lost}$ .

$1680\ \text{J}$ (correct answer)
$2320\ \text{J}$
$320\ \text{J}$
$620\ \text{J}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this device, the input energy is E_in = 2000 J and the useful output energy is E_out = 320 J, so efficiency η = (320/2000) × 100% = 16%. The remaining energy E_waste = 2000 - 320 = 1680 J is lost as thermal energy, which is characteristic of solar panels where incomplete absorption and conversion losses cause heating. Choice A is correct because it correctly identifies waste energy as E_in - E_out. Choice B incorrectly adds energies instead of subtracting to find waste. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 5

A battery charging system stores $E_{out}=760\ \text{J}$ of chemical energy while $E_{in}=800\ \text{J}$ of electrical energy is supplied. What fraction of the input energy is wasted (lost as heat), expressed as a percentage of the input?

$5\%$ (correct answer)
$40\%$
$95\%$
$105\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this device, the input energy is E_in = 800 J and the useful output energy is E_out = 760 J, so efficiency η = (760/800) × 100% = 95%. The remaining energy E_waste = 800 - 760 = 40 J is lost as thermal energy, which is characteristic of battery chargers where electrical resistance causes heating, and the waste fraction is (40/800) × 100% = 5%. Choice A is correct because it properly calculates the waste fraction as ((input - output)/input) × 100%. Choice C incorrectly states the efficiency instead of the waste percentage. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 6

An electric motor takes in electrical power $P_{in} = 500 \, \text{W}$ and delivers useful mechanical power $P_{out} = 400 \, \text{W}$ . The rest is lost as thermal energy in the motor. What is the motor’s efficiency (as a percentage)?

Use $\eta = \dfrac{P_{out}}{P_{in}} \times 100\%$

$125\%$
$0.80\%$
$80\%$ (correct answer)
$20\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as $η = \dfrac{\text{useful energy output}}{\text{total energy input}} \times 100\%$ , and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved ( $E_{\text{input}} = E_{\text{useful}} + E_{\text{waste}}$ ), but not all input energy converts to the desired useful form. For this device, the input power is $P_{\text{in}} = 500 \ \text{W}$ and the useful output power is $P_{\text{out}} = 400 \ \text{W}$ , so efficiency $η = \dfrac{400}{500} \times 100\% = 80\%$ . The remaining power $P_{\text{waste}} = 500 - 400 = 100 \ \text{W}$ is lost as thermal energy, which is characteristic of electric motors where electrical resistance and friction cause heating. Choice C is correct because it properly calculates efficiency as $(\text{output}/\text{input}) \times 100\%$ . Choice A inverts the ratio calculating input/output instead of output/input, giving $>100\%$ which violates conservation. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute $η = (\text{output}/\text{input}) \times 100\%$ , (4) verify $η$ between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 7

Two light bulbs convert electrical power into light (useful) and heat (waste). Bulb 1 has $P_{in}=100\ \text{W}$ and produces $P_{light}=5\ \text{W}$ . Bulb 2 has $P_{in}=20\ \text{W}$ and produces $P_{light}=8\ \text{W}$ . Which bulb is more efficient at producing light, and by how many percentage points?

Use: $\eta=\left(\frac{P_{light}}{P_{in}}\right)\times 100\%$

Bulb 1 by $35$ percentage points
Bulb 2 by $35$ percentage points (correct answer)
Bulb 2 by $5$ percentage points
Bulb 1 by $5$ percentage points

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For Bulb 1: η₁ = (P_light/P_in) × 100% = (5/100) × 100% = 5%; For Bulb 2: η₂ = (P_light/P_in) × 100% = (8/20) × 100% = 40%. The difference is η₂ - η₁ = 40% - 5% = 35 percentage points, with Bulb 2 being more efficient. The remaining power is lost as thermal energy, which is characteristic of incandescent bulbs where electrical resistance in the filament produces mostly heat rather than light. Choice B is correct because it properly identifies Bulb 2 as more efficient by 35 percentage points (40% - 5% = 35%). Choice A incorrectly identifies Bulb 1 as more efficient; Choice C calculates the wrong difference; Choice D also calculates the wrong difference. To calculate efficiency: (1) compute each bulb's efficiency separately, (2) compare values to find which is higher, (3) subtract to find difference in percentage points. Higher efficiency means more input converts to useful light output, less wasted as heat.

Question 8

A heat engine takes in thermal energy $E_{in}=5000\ \text{J}$ from fuel and produces useful mechanical work $E_{out}=1500\ \text{J}$ . What is the engine efficiency $\eta$ (as a percentage)?

Use: $\eta = \left(\frac{E_{out}}{E_{in}}\right)\times 100\%$

$30\%$ (correct answer)
$3.0\%$
$70\%$
$333\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this heat engine, the input energy is E_in = 5000 J and the useful output energy is E_out = 1500 J, so efficiency η = (E_out/E_in) × 100% = (1500/5000) × 100% = 0.30 × 100% = 30%. The remaining energy E_waste = E_in - E_out = 5000 - 1500 = 3500 J is lost as waste heat, which is characteristic of heat engines where incomplete combustion and exhaust gases carry away thermal energy. Choice A is correct because it properly calculates efficiency as (1500/5000) × 100% = 30%. Choice B (3.0%) makes a decimal place error; Choice C (70%) calculates the waste percentage instead of efficiency; Choice D (333%) inverts the ratio calculating input/output instead of output/input, giving >100% which violates conservation. To calculate efficiency: (1) identify useful output energy, (2) identify total input energy, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted.

Question 9

An incandescent light bulb uses $P_{in}=100\,\text{W}$ of electrical power and produces $P_{out}=5\,\text{W}$ of useful light. The rest is released as heat. How much power is lost as thermal energy, $P_{lost}$ ?

$105\,\text{W}$
$95\,\text{W}$ (correct answer)
$20\,\text{W}$
$5\,\text{W}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this device, the input power is P_in = 100 W and the useful output power is P_out = 5 W, so efficiency η = (P_out/P_in) × 100% = (5/100) × 100% = 5%. The remaining power P_waste = 100 - 5 = 95 W is lost as thermal energy, which is characteristic of incandescent bulbs where electrical resistance in the filament causes heating. Choice B is correct because it correctly identifies waste energy as P_in - P_out. Choice A is wrong because it incorrectly adds energies instead of subtracting to find waste. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 10

A battery charger transfers $E_{in}=1500\,\text{J}$ of electrical energy to a rechargeable battery. After charging, the battery has stored $E_{out}=1200\,\text{J}$ of chemical energy. The rest is lost mainly as heat. How much energy is lost, $E_{lost}$ ?

$300\,\text{J}$ (correct answer)
$2700\,\text{J}$
$1200\,\text{J}$
$80\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this device, the input energy is E_in = 1500 J and the useful output energy is E_out = 1200 J, so efficiency η = (E_out/E_in) × 100% = (1200/1500) × 100% = 80%. The remaining energy E_waste = 1500 - 1200 = 300 J is lost as thermal energy, which is characteristic of battery chargers where electrical resistance causes heating. Choice A is correct because it correctly identifies waste energy as E_in - E_out. Choice B is wrong because it incorrectly adds energies instead of subtracting to find waste. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 11

A device converts energy with $E_{in}=3600\,\text{J}$ and produces useful output $E_{out}=900\,\text{J}$ . The remainder is waste (mostly thermal). What percentage of the input energy is wasted?

$25\%$
$75\%$ (correct answer)
$400\%$
$0.75\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this device, the input energy is E_in = 3600 J and the useful output energy is E_out = 900 J, so efficiency η = (900/3600) × 100% = 25%, and wasted percentage = 100% - 25% = 75%. The remaining energy E_waste = 3600 - 900 = 2700 J is lost as thermal energy, which is characteristic of energy conversion devices where resistance or friction causes losses. Choice B is correct because it correctly identifies wasted percentage as 100% minus efficiency. Choice C is wrong because it inverts the ratio calculating input/output instead of output/input, giving >100% which violates conservation. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 12

A battery charging system receives $E_{in}=4000\ \text{J}$ of electrical energy. Measurements show $E_{lost}=600\ \text{J}$ is dissipated as heat. How much useful chemical energy is stored in the battery, $E_{out}$ ?

Use: $E_{in}=E_{out}+E_{lost}$

$4600\ \text{J}$
$3400\ \text{J}$ (correct answer)
$2400\ \text{J}$
$0.85\ \text{J}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this battery charging system, the input energy is E_in = 4000 J and the waste energy is E_lost = 600 J, so using E_in = E_out + E_lost, we get E_out = E_in - E_lost = 4000 - 600 = 3400 J. The remaining energy E_waste = 600 J is lost as thermal energy, which is characteristic of battery chargers where electrical resistance and chemical reactions produce heat. Choice B is correct because it properly calculates useful energy stored as E_out = E_in - E_lost = 4000 - 600 = 3400 J. Choice A (4600 J) incorrectly adds energies instead of subtracting; Choice C (2400 J) makes an arithmetic error; Choice D (0.85 J) appears to calculate efficiency as a decimal but gives energy units. To calculate efficiency: (1) identify total input energy, (2) identify waste energy, (3) compute useful output = input - waste, (4) verify output is positive and less than input. Higher efficiency means more input converts to useful chemical energy, less wasted as heat.

Question 13

A solar panel receives $E_{in}=2000\ \text{J}$ of solar radiation in a short time interval and produces $E_{out}=320\ \text{J}$ of useful electrical energy. How much energy is lost as waste (mostly thermal), $E_{lost}$ ?

Use: $E_{in}=E_{out}+E_{lost}$

$1680\ \text{J}$ (correct answer)
$2320\ \text{J}$
$640\ \text{J}$
$620\ \text{J}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this solar panel, the input energy is E_in = 2000 J and the useful output energy is E_out = 320 J, so the waste energy E_lost = E_in - E_out = 2000 - 320 = 1680 J. The remaining energy E_waste = 1680 J is lost as thermal energy, which is characteristic of solar panels where photons not converted to electricity heat the panel material. Choice A is correct because it properly calculates waste energy as E_in - E_out = 2000 - 320 = 1680 J. Choice B (2320 J) incorrectly adds energies instead of subtracting; Choice C (640 J) makes an arithmetic error; Choice D (620 J) also makes an arithmetic error. To calculate efficiency: (1) identify useful output energy, (2) identify total input energy, (3) compute waste = input - output, (4) verify waste is positive and less than input. Higher efficiency means more input converts to useful output, less wasted.

Question 14

A light bulb uses $E_{in}=1500\ \text{J}$ of electrical energy in a short time and produces $E_{light}=300\ \text{J}$ of useful light energy. What is the efficiency $\eta$ (as a percentage)?

$500\%$
$80\%$
$20\%$ (correct answer)
$0.20\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this light bulb, the input energy is E_in = 1500 J and the useful output energy is E_light = 300 J, so efficiency η = (E_light/E_in) × 100% = (300/1500) × 100% = 0.20 × 100% = 20%. The remaining energy E_waste = E_in - E_out = 1500 - 300 = 1200 J is lost as thermal energy, which is characteristic of incandescent bulbs where electrical resistance in the filament produces mostly heat rather than light. Choice C is correct because it properly calculates efficiency as (300/1500) × 100% = 20%. Choice A (500%) inverts the ratio calculating input/output instead of output/input, giving >100% which violates conservation; Choice B (80%) calculates the waste percentage instead of efficiency; Choice D (0.20%) forgets to multiply by 100% to convert decimal to percentage. To calculate efficiency: (1) identify useful output energy, (2) identify total input energy, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted.

Question 15

A solar panel produces $P_{out}=90\ \text{W}$ of electrical power from $P_{in}=600\ \text{W}$ of incoming solar power. What percentage of the input power is lost as waste (mostly thermal)?

Use: $P_{lost}=P_{in}-P_{out}$ and $\%\text{ lost}=\left(\frac{P_{lost}}{P_{in}}\right)\times 100\%$

$15\%$
$85\%$ (correct answer)
$90\%$
$115\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this solar panel, P_in = 600 W and P_out = 90 W, so P_lost = P_in - P_out = 600 - 90 = 510 W, and the percentage lost = (P_lost/P_in) × 100% = (510/600) × 100% = 0.85 × 100% = 85%. The remaining energy (510 W) is lost as thermal energy, which is characteristic of solar panels where photons not converted to electricity heat the panel material. Choice B is correct because it properly calculates waste percentage as (510/600) × 100% = 85%. Choice A (15%) calculates the efficiency instead of waste percentage; Choice C (90%) makes an arithmetic error; Choice D (115%) gives >100% which violates conservation. To calculate efficiency: (1) find waste power = input - output, (2) compute waste percentage = (waste/input) × 100%, (3) verify waste percentage + efficiency = 100%. Higher efficiency means less input wasted as heat.

Question 16

A solar panel operates at an efficiency of $18\%$ . If it receives $E_{in}=3000\ \text{J}$ of solar radiation, how much useful electrical energy $E_{out}$ does it produce?

Use: $E_{out}=\eta E_{in}$ with $\eta$ as a decimal.

$540\ \text{J}$ (correct answer)
$166.7\ \text{J}$
$3000\ \text{J}$
$5400\ \text{J}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this solar panel operating at 18% efficiency with input energy E_in = 3000 J, the useful output energy E_out = η × E_in = 0.18 × 3000 = 540 J (using η = 18% = 0.18 as decimal). The remaining energy E_waste = E_in - E_out = 3000 - 540 = 2460 J is lost as thermal energy, which is characteristic of solar panels where photons not converted to electricity heat the panel. Choice A is correct because it properly calculates output energy as E_out = η × E_in = 0.18 × 3000 = 540 J. Choice B (166.7 J) divides instead of multiplying (3000/18); Choice C (3000 J) would require 100% efficiency which is impossible; Choice D (5400 J) uses 180% instead of 18% efficiency. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. When given efficiency, output = efficiency × input (with efficiency as decimal).

Question 17

A heat engine produces $P_{out}=300\ \text{W}$ of useful mechanical power at an efficiency of $30\%$ . What input thermal power $P_{in}$ is required?

Use: $\eta = \left(\frac{P_{out}}{P_{in}}\right)\times 100\%$ .

$100\ \text{W}$
$900\ \text{W}$
$1000\ \text{W}$ (correct answer)
$0.90\ \text{W}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this heat engine with output power P_out = 300 W and efficiency η = 30% = 0.30, we can rearrange η = P_out/P_in to get P_in = P_out/η = 300/0.30 = 1000 W. The remaining power P_waste = P_in - P_out = 1000 - 300 = 700 W is expelled as waste heat, which is characteristic of heat engines where thermodynamic limits prevent complete conversion of thermal energy to mechanical work. Choice C is correct because it properly calculates input power as P_in = P_out/η = 300/0.30 = 1000 W. Choice A (100 W) makes arithmetic error; Choice B (900 W) calculates waste power instead of input power; Choice D (0.90 W) appears to use wrong calculation. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. When given efficiency and output, input = output/efficiency (with efficiency as decimal).

Question 18

A solar panel receives $E_{in} = 2000\ \text{J}$ of solar radiation energy in a short time interval and produces $E_{out} = 320\ \text{J}$ of useful electrical energy. Assuming the rest becomes thermal energy, what is the panel's efficiency $\eta$ (in %)?

$16\%$ (correct answer)
$6.25\%$
$84\%$
$0.16\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this solar panel, the input energy is E_in = 2000 J and the useful output energy is E_out = 320 J, so efficiency η = (E_out/E_in) × 100% = (320/2000) × 100% = 0.16 × 100% = 16%. The remaining energy E_waste = E_in - E_out = 2000 - 320 = 1680 J is lost as thermal energy, which is characteristic of solar panels where photons not converted to electricity heat the panel material. Choice A is correct because it properly calculates efficiency as (output/input) × 100% = (320/2000) × 100% = 16%. Choice B (6.25%) makes arithmetic error dividing 2000/320 instead of 320/2000; Choice C (84%) incorrectly calculates waste percentage (1680/2000) instead of efficiency; Choice D (0.16%) forgets to multiply by 100% to convert decimal to percentage. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Solar panels typically have 15-20% efficiency due to fundamental limits in photon-to-electron conversion.

Question 19

A battery charger supplies $E_{in} = 1500\ \text{J}$ of electrical energy to a rechargeable battery. The battery stores $E_{out} = 1200\ \text{J}$ as chemical energy. How much energy is lost as waste (mostly thermal), $E_{lost}$ ?

Use: $E_{in} = E_{out} + E_{lost}$ .

$2700\ \text{J}$
$300\ \text{J}$ (correct answer)
$1800\ \text{J}$
$0.80\ \text{J}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this battery charger, the input energy is E_in = 1500 J and the useful output energy stored is E_out = 1200 J, so the waste energy E_lost = E_in - E_out = 1500 - 1200 = 300 J. The remaining energy E_waste = 300 J is lost as thermal energy, which is characteristic of battery chargers where electrical resistance and chemical conversion inefficiencies generate heat. Choice B is correct because it properly calculates waste energy as E_in - E_out = 1500 - 1200 = 300 J. Choice A (2700 J) incorrectly adds energies instead of subtracting; Choice C (1800 J) makes arithmetic error; Choice D (0.80 J) appears to calculate efficiency as decimal (0.80) instead of energy loss. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Battery charging efficiency (80% here) is limited by internal resistance and electrochemical conversion losses.

Question 20

Two light bulbs are tested for a fixed time interval.

Bulb 1 (incandescent): $P_{in}=100\ \text{W}$ electrical, useful light output $P_{out}=5\ \text{W}$ .
Bulb 2 (LED): $P_{in}=20\ \text{W}$ electrical, useful light output $P_{out}=8\ \text{W}$ .

Which bulb is more efficient at converting electrical power into useful light, and what is that efficiency?

Bulb 1, $5\%$
Bulb 2, $40\%$ (correct answer)
Bulb 1, $20\%$
Bulb 2, $2.5\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For Bulb 1 (incandescent), efficiency η₁ = (P_out/P_in) × 100% = (5/100) × 100% = 5%; for Bulb 2 (LED), efficiency η₂ = (P_out/P_in) × 100% = (8/20) × 100% = 40%. The remaining power is lost as thermal energy, which is characteristic of light bulbs where electrical resistance generates heat—incandescent bulbs waste 95% as heat while LEDs waste only 60%. Choice B is correct because it correctly identifies Bulb 2 (LED) as more efficient and properly calculates its efficiency as (8/20) × 100% = 40%. Choice A (Bulb 1, 5%) correctly calculates incandescent efficiency but wrongly identifies it as more efficient; Choice C (Bulb 1, 20%) makes arithmetic error; Choice D (Bulb 2, 2.5%) inverts the ratio. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. LEDs are far more efficient than incandescent bulbs because they produce light through electroluminescence rather than heating a filament.

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Physics Quiz

Physics Quiz: Evaluate Device Energy Efficiency

Practice Evaluate Device Energy Efficiency in Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Evaluate Device Energy Efficiency, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$30\%$ (correct answer)
$70\%$
$3.0\%$
$333\%$

Question 2

Two light bulbs produce light (useful) and heat (waste).

Incandescent bulb: $P_{in}=100\ \text{W}$ electrical, $P_{out}=5\ \text{W}$ light.
LED bulb: $P_{in}=20\ \text{W}$ electrical, $P_{out}=8\ \text{W}$ light.

Which bulb is more efficient, and by how many percentage points (LED efficiency minus incandescent efficiency)?

Incandescent by $35$ percentage points
LED by $35$ percentage points (correct answer)
LED by $3.5$ percentage points
Incandescent by $3.5$ percentage points

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For the incandescent bulb, the input power is P_in = 100 W and the useful output power is P_out = 5 W, so efficiency η = (5/100) × 100% = 5%; for the LED, P_in = 20 W and P_out = 8 W, so η = (8/20) × 100% = 40%. The difference is 40% - 5% = 35 percentage points, with the LED being more efficient, characteristic of bulbs where incandescent loses more to heat via filament resistance while LED is more direct in light conversion. Choice B is correct because it accurately compares efficiencies showing which device converts larger fraction. Choice A incorrectly identifies the incandescent as more efficient. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 3

A heat engine takes in $E_{in}=5000\ \text{J}$ of thermal energy from fuel and produces $E_{out}=1400\ \text{J}$ of useful mechanical work. What is the efficiency of the engine (as a percentage)?

$3.6\%$
$28\%$ (correct answer)
$72\%$
$357\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this device, the input energy is E_in = 5000 J and the useful output energy is E_out = 1400 J, so efficiency η = (1400/5000) × 100% = 28%. The remaining energy E_waste = 5000 - 1400 = 3600 J is lost as thermal energy, which is characteristic of heat engines where incomplete combustion and thermodynamic limits cause losses. Choice B is correct because it properly calculates efficiency as (output/input) × 100%. Choice D inverts the ratio calculating input/output instead of output/input, giving >100% which violates conservation. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 4

Use $E_{in}=E_{out}+E_{lost}$ .

$1680\ \text{J}$ (correct answer)
$2320\ \text{J}$
$320\ \text{J}$
$620\ \text{J}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this device, the input energy is E_in = 2000 J and the useful output energy is E_out = 320 J, so efficiency η = (320/2000) × 100% = 16%. The remaining energy E_waste = 2000 - 320 = 1680 J is lost as thermal energy, which is characteristic of solar panels where incomplete absorption and conversion losses cause heating. Choice A is correct because it correctly identifies waste energy as E_in - E_out. Choice B incorrectly adds energies instead of subtracting to find waste. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 5

$5\%$ (correct answer)
$40\%$
$95\%$
$105\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this device, the input energy is E_in = 800 J and the useful output energy is E_out = 760 J, so efficiency η = (760/800) × 100% = 95%. The remaining energy E_waste = 800 - 760 = 40 J is lost as thermal energy, which is characteristic of battery chargers where electrical resistance causes heating, and the waste fraction is (40/800) × 100% = 5%. Choice A is correct because it properly calculates the waste fraction as ((input - output)/input) × 100%. Choice C incorrectly states the efficiency instead of the waste percentage. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 6

Use $\eta = \dfrac{P_{out}}{P_{in}} \times 100\%$

$125\%$
$0.80\%$
$80\%$ (correct answer)
$20\%$

Question 7

Use: $\eta=\left(\frac{P_{light}}{P_{in}}\right)\times 100\%$

Bulb 1 by $35$ percentage points
Bulb 2 by $35$ percentage points (correct answer)
Bulb 2 by $5$ percentage points
Bulb 1 by $5$ percentage points

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For Bulb 1: η₁ = (P_light/P_in) × 100% = (5/100) × 100% = 5%; For Bulb 2: η₂ = (P_light/P_in) × 100% = (8/20) × 100% = 40%. The difference is η₂ - η₁ = 40% - 5% = 35 percentage points, with Bulb 2 being more efficient. The remaining power is lost as thermal energy, which is characteristic of incandescent bulbs where electrical resistance in the filament produces mostly heat rather than light. Choice B is correct because it properly identifies Bulb 2 as more efficient by 35 percentage points (40% - 5% = 35%). Choice A incorrectly identifies Bulb 1 as more efficient; Choice C calculates the wrong difference; Choice D also calculates the wrong difference. To calculate efficiency: (1) compute each bulb's efficiency separately, (2) compare values to find which is higher, (3) subtract to find difference in percentage points. Higher efficiency means more input converts to useful light output, less wasted as heat.

Question 8

A heat engine takes in thermal energy $E_{in}=5000\ \text{J}$ from fuel and produces useful mechanical work $E_{out}=1500\ \text{J}$ . What is the engine efficiency $\eta$ (as a percentage)?

Use: $\eta = \left(\frac{E_{out}}{E_{in}}\right)\times 100\%$

$30\%$ (correct answer)
$3.0\%$
$70\%$
$333\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this heat engine, the input energy is E_in = 5000 J and the useful output energy is E_out = 1500 J, so efficiency η = (E_out/E_in) × 100% = (1500/5000) × 100% = 0.30 × 100% = 30%. The remaining energy E_waste = E_in - E_out = 5000 - 1500 = 3500 J is lost as waste heat, which is characteristic of heat engines where incomplete combustion and exhaust gases carry away thermal energy. Choice A is correct because it properly calculates efficiency as (1500/5000) × 100% = 30%. Choice B (3.0%) makes a decimal place error; Choice C (70%) calculates the waste percentage instead of efficiency; Choice D (333%) inverts the ratio calculating input/output instead of output/input, giving >100% which violates conservation. To calculate efficiency: (1) identify useful output energy, (2) identify total input energy, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted.

Question 9

$105\,\text{W}$
$95\,\text{W}$ (correct answer)
$20\,\text{W}$
$5\,\text{W}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this device, the input power is P_in = 100 W and the useful output power is P_out = 5 W, so efficiency η = (P_out/P_in) × 100% = (5/100) × 100% = 5%. The remaining power P_waste = 100 - 5 = 95 W is lost as thermal energy, which is characteristic of incandescent bulbs where electrical resistance in the filament causes heating. Choice B is correct because it correctly identifies waste energy as P_in - P_out. Choice A is wrong because it incorrectly adds energies instead of subtracting to find waste. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 10

$300\,\text{J}$ (correct answer)
$2700\,\text{J}$
$1200\,\text{J}$
$80\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this device, the input energy is E_in = 1500 J and the useful output energy is E_out = 1200 J, so efficiency η = (E_out/E_in) × 100% = (1200/1500) × 100% = 80%. The remaining energy E_waste = 1500 - 1200 = 300 J is lost as thermal energy, which is characteristic of battery chargers where electrical resistance causes heating. Choice A is correct because it correctly identifies waste energy as E_in - E_out. Choice B is wrong because it incorrectly adds energies instead of subtracting to find waste. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 11

A device converts energy with $E_{in}=3600\,\text{J}$ and produces useful output $E_{out}=900\,\text{J}$ . The remainder is waste (mostly thermal). What percentage of the input energy is wasted?

$25\%$
$75\%$ (correct answer)
$400\%$
$0.75\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this device, the input energy is E_in = 3600 J and the useful output energy is E_out = 900 J, so efficiency η = (900/3600) × 100% = 25%, and wasted percentage = 100% - 25% = 75%. The remaining energy E_waste = 3600 - 900 = 2700 J is lost as thermal energy, which is characteristic of energy conversion devices where resistance or friction causes losses. Choice B is correct because it correctly identifies wasted percentage as 100% minus efficiency. Choice C is wrong because it inverts the ratio calculating input/output instead of output/input, giving >100% which violates conservation. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

Question 12

Use: $E_{in}=E_{out}+E_{lost}$

$4600\ \text{J}$
$3400\ \text{J}$ (correct answer)
$2400\ \text{J}$
$0.85\ \text{J}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this battery charging system, the input energy is E_in = 4000 J and the waste energy is E_lost = 600 J, so using E_in = E_out + E_lost, we get E_out = E_in - E_lost = 4000 - 600 = 3400 J. The remaining energy E_waste = 600 J is lost as thermal energy, which is characteristic of battery chargers where electrical resistance and chemical reactions produce heat. Choice B is correct because it properly calculates useful energy stored as E_out = E_in - E_lost = 4000 - 600 = 3400 J. Choice A (4600 J) incorrectly adds energies instead of subtracting; Choice C (2400 J) makes an arithmetic error; Choice D (0.85 J) appears to calculate efficiency as a decimal but gives energy units. To calculate efficiency: (1) identify total input energy, (2) identify waste energy, (3) compute useful output = input - waste, (4) verify output is positive and less than input. Higher efficiency means more input converts to useful chemical energy, less wasted as heat.

Question 13

Use: $E_{in}=E_{out}+E_{lost}$

$1680\ \text{J}$ (correct answer)
$2320\ \text{J}$
$640\ \text{J}$
$620\ \text{J}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this solar panel, the input energy is E_in = 2000 J and the useful output energy is E_out = 320 J, so the waste energy E_lost = E_in - E_out = 2000 - 320 = 1680 J. The remaining energy E_waste = 1680 J is lost as thermal energy, which is characteristic of solar panels where photons not converted to electricity heat the panel material. Choice A is correct because it properly calculates waste energy as E_in - E_out = 2000 - 320 = 1680 J. Choice B (2320 J) incorrectly adds energies instead of subtracting; Choice C (640 J) makes an arithmetic error; Choice D (620 J) also makes an arithmetic error. To calculate efficiency: (1) identify useful output energy, (2) identify total input energy, (3) compute waste = input - output, (4) verify waste is positive and less than input. Higher efficiency means more input converts to useful output, less wasted.

Question 14

A light bulb uses $E_{in}=1500\ \text{J}$ of electrical energy in a short time and produces $E_{light}=300\ \text{J}$ of useful light energy. What is the efficiency $\eta$ (as a percentage)?

$500\%$
$80\%$
$20\%$ (correct answer)
$0.20\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this light bulb, the input energy is E_in = 1500 J and the useful output energy is E_light = 300 J, so efficiency η = (E_light/E_in) × 100% = (300/1500) × 100% = 0.20 × 100% = 20%. The remaining energy E_waste = E_in - E_out = 1500 - 300 = 1200 J is lost as thermal energy, which is characteristic of incandescent bulbs where electrical resistance in the filament produces mostly heat rather than light. Choice C is correct because it properly calculates efficiency as (300/1500) × 100% = 20%. Choice A (500%) inverts the ratio calculating input/output instead of output/input, giving >100% which violates conservation; Choice B (80%) calculates the waste percentage instead of efficiency; Choice D (0.20%) forgets to multiply by 100% to convert decimal to percentage. To calculate efficiency: (1) identify useful output energy, (2) identify total input energy, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Higher efficiency means more input converts to useful output, less wasted.

Question 15

A solar panel produces $P_{out}=90\ \text{W}$ of electrical power from $P_{in}=600\ \text{W}$ of incoming solar power. What percentage of the input power is lost as waste (mostly thermal)?

Use: $P_{lost}=P_{in}-P_{out}$ and $\%\text{ lost}=\left(\frac{P_{lost}}{P_{in}}\right)\times 100\%$

$15\%$
$85\%$ (correct answer)
$90\%$
$115\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this solar panel, P_in = 600 W and P_out = 90 W, so P_lost = P_in - P_out = 600 - 90 = 510 W, and the percentage lost = (P_lost/P_in) × 100% = (510/600) × 100% = 0.85 × 100% = 85%. The remaining energy (510 W) is lost as thermal energy, which is characteristic of solar panels where photons not converted to electricity heat the panel material. Choice B is correct because it properly calculates waste percentage as (510/600) × 100% = 85%. Choice A (15%) calculates the efficiency instead of waste percentage; Choice C (90%) makes an arithmetic error; Choice D (115%) gives >100% which violates conservation. To calculate efficiency: (1) find waste power = input - output, (2) compute waste percentage = (waste/input) × 100%, (3) verify waste percentage + efficiency = 100%. Higher efficiency means less input wasted as heat.

Question 16

A solar panel operates at an efficiency of $18\%$ . If it receives $E_{in}=3000\ \text{J}$ of solar radiation, how much useful electrical energy $E_{out}$ does it produce?

Use: $E_{out}=\eta E_{in}$ with $\eta$ as a decimal.

$540\ \text{J}$ (correct answer)
$166.7\ \text{J}$
$3000\ \text{J}$
$5400\ \text{J}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this solar panel operating at 18% efficiency with input energy E_in = 3000 J, the useful output energy E_out = η × E_in = 0.18 × 3000 = 540 J (using η = 18% = 0.18 as decimal). The remaining energy E_waste = E_in - E_out = 3000 - 540 = 2460 J is lost as thermal energy, which is characteristic of solar panels where photons not converted to electricity heat the panel. Choice A is correct because it properly calculates output energy as E_out = η × E_in = 0.18 × 3000 = 540 J. Choice B (166.7 J) divides instead of multiplying (3000/18); Choice C (3000 J) would require 100% efficiency which is impossible; Choice D (5400 J) uses 180% instead of 18% efficiency. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. When given efficiency, output = efficiency × input (with efficiency as decimal).

Question 17

A heat engine produces $P_{out}=300\ \text{W}$ of useful mechanical power at an efficiency of $30\%$ . What input thermal power $P_{in}$ is required?

Use: $\eta = \left(\frac{P_{out}}{P_{in}}\right)\times 100\%$ .

$100\ \text{W}$
$900\ \text{W}$
$1000\ \text{W}$ (correct answer)
$0.90\ \text{W}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this heat engine with output power P_out = 300 W and efficiency η = 30% = 0.30, we can rearrange η = P_out/P_in to get P_in = P_out/η = 300/0.30 = 1000 W. The remaining power P_waste = P_in - P_out = 1000 - 300 = 700 W is expelled as waste heat, which is characteristic of heat engines where thermodynamic limits prevent complete conversion of thermal energy to mechanical work. Choice C is correct because it properly calculates input power as P_in = P_out/η = 300/0.30 = 1000 W. Choice A (100 W) makes arithmetic error; Choice B (900 W) calculates waste power instead of input power; Choice D (0.90 W) appears to use wrong calculation. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. When given efficiency and output, input = output/efficiency (with efficiency as decimal).

Question 18

$16\%$ (correct answer)
$6.25\%$
$84\%$
$0.16\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this solar panel, the input energy is E_in = 2000 J and the useful output energy is E_out = 320 J, so efficiency η = (E_out/E_in) × 100% = (320/2000) × 100% = 0.16 × 100% = 16%. The remaining energy E_waste = E_in - E_out = 2000 - 320 = 1680 J is lost as thermal energy, which is characteristic of solar panels where photons not converted to electricity heat the panel material. Choice A is correct because it properly calculates efficiency as (output/input) × 100% = (320/2000) × 100% = 16%. Choice B (6.25%) makes arithmetic error dividing 2000/320 instead of 320/2000; Choice C (84%) incorrectly calculates waste percentage (1680/2000) instead of efficiency; Choice D (0.16%) forgets to multiply by 100% to convert decimal to percentage. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Solar panels typically have 15-20% efficiency due to fundamental limits in photon-to-electron conversion.

Question 19

Use: $E_{in} = E_{out} + E_{lost}$ .

$2700\ \text{J}$
$300\ \text{J}$ (correct answer)
$1800\ \text{J}$
$0.80\ \text{J}$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For this battery charger, the input energy is E_in = 1500 J and the useful output energy stored is E_out = 1200 J, so the waste energy E_lost = E_in - E_out = 1500 - 1200 = 300 J. The remaining energy E_waste = 300 J is lost as thermal energy, which is characteristic of battery chargers where electrical resistance and chemical conversion inefficiencies generate heat. Choice B is correct because it properly calculates waste energy as E_in - E_out = 1500 - 1200 = 300 J. Choice A (2700 J) incorrectly adds energies instead of subtracting; Choice C (1800 J) makes arithmetic error; Choice D (0.80 J) appears to calculate efficiency as decimal (0.80) instead of energy loss. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. Battery charging efficiency (80% here) is limited by internal resistance and electrochemical conversion losses.

Question 20

Two light bulbs are tested for a fixed time interval.

Bulb 1 (incandescent): $P_{in}=100\ \text{W}$ electrical, useful light output $P_{out}=5\ \text{W}$ .
Bulb 2 (LED): $P_{in}=20\ \text{W}$ electrical, useful light output $P_{out}=8\ \text{W}$ .

Which bulb is more efficient at converting electrical power into useful light, and what is that efficiency?

Bulb 1, $5\%$
Bulb 2, $40\%$ (correct answer)
Bulb 1, $20\%$
Bulb 2, $2.5\%$

Explanation: This question tests understanding of energy efficiency in devices that convert energy from one form to another. Efficiency is defined as η = (useful energy output / total energy input) × 100%, and no real device achieves 100% efficiency because some energy is always lost (usually as waste heat) due to friction, electrical resistance, or incomplete combustion—energy is conserved (E_input = E_useful + E_waste), but not all input energy converts to the desired useful form. For Bulb 1 (incandescent), efficiency η₁ = (P_out/P_in) × 100% = (5/100) × 100% = 5%; for Bulb 2 (LED), efficiency η₂ = (P_out/P_in) × 100% = (8/20) × 100% = 40%. The remaining power is lost as thermal energy, which is characteristic of light bulbs where electrical resistance generates heat—incandescent bulbs waste 95% as heat while LEDs waste only 60%. Choice B is correct because it correctly identifies Bulb 2 (LED) as more efficient and properly calculates its efficiency as (8/20) × 100% = 40%. Choice A (Bulb 1, 5%) correctly calculates incandescent efficiency but wrongly identifies it as more efficient; Choice C (Bulb 1, 20%) makes arithmetic error; Choice D (Bulb 2, 2.5%) inverts the ratio. To calculate efficiency: (1) identify useful output energy/power, (2) identify total input energy/power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste energy = input - output. LEDs are far more efficient than incandescent bulbs because they produce light through electroluminescence rather than heating a filament.