All questions
Question 1
A solar panel converts 18% of incoming solar energy into electrical energy. If Ein=2500 J of solar energy strikes the panel, how much electrical energy is produced, Eout?
- 450 J (correct answer)
- 2000 J
- 318 J
- 13889 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 = 2500 J and efficiency η = 18% = 0.18, so useful output energy E_out = 0.18 × 2500 = 450 J. The remaining energy E_waste = 2500 - 450 = 2050 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 properly calculates output as η × input. Choice D inverts the ratio, calculating input / η which is incorrect for output. 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
A heat engine takes in thermal power Pin=900W from burning fuel and produces useful mechanical power Pout=270W. The rest is waste heat to the surroundings. What is the engine's efficiency (as a percentage)?
- 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 3
An electric motor has efficiency η=80% and delivers useful mechanical power Pout=320 W. What electrical input power Pin is required? (Use η=(Pout/Pin)×100%.)
- 256 W
- 400 W (correct answer)
- 640 W
- 0.80 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 motor with efficiency η = 80% and output power P_out = 320 W, we need to find input power: η = (P_out/P_in) × 100%, so 80 = (320/P_in) × 100, which gives P_in = 320/(0.80) = 400 W. The remaining power P_waste = P_in - P_out = 80 W is lost as thermal energy, which is characteristic of electric motors where electrical resistance and friction cause heating. Choice B is correct because it properly calculates input power from η = (P_out/P_in) × 100%, rearranging to P_in = P_out/(η/100) = 320/0.80 = 400 W. Choice A incorrectly multiplies 320 × 0.80 = 256 W; Choice C inverts the efficiency calculation; Choice D makes an arithmetic error. 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.
Question 4
A battery charger takes in Pin=120 W from the wall and stores useful chemical energy in the battery at a rate of Pout=102 W. What is the efficiency η (in %)?
- 85% (correct answer)
- 118%
- 0.85%
- 15%
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 power is P_in = 120 W and the useful output power is P_out = 102 W, so efficiency η = (P_out/P_in) × 100% = (102/120) × 100% = 85%. The remaining power P_waste = P_in - P_out = 18 W is lost as heat, which is characteristic of battery chargers where electrical resistance and chemical conversion inefficiencies generate heat. Choice A is correct because it properly calculates efficiency as (output/input) × 100% = (102/120) × 100% = 85%. Choice B incorrectly inverts the ratio calculating input/output instead of output/input, giving 118% which violates conservation; Choice C forgets to multiply by 100% to convert decimal to percentage; Choice D calculates the waste percentage (18/120 = 15%) instead of 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. Higher efficiency means more input converts to useful output, less wasted.
Question 5
An electric motor has efficiency η=85% and receives input power Pin=200W. How much useful mechanical output power Pout does it deliver?
- 170W (correct answer)
- 235W
- 15W
- 85W
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 efficiency is η = 85% and input power is P_in = 200 W, so useful output power P_out = (η/100%) × P_in = 0.85 × 200 = 170 W. The remaining power P_waste = 200 - 170 = 30 W is lost as thermal energy, which is characteristic of electric motors where electrical resistance in windings causes heating. Choice A is correct because it properly calculates output as efficiency times input. Choice B is wrong because it makes arithmetic error in multiplication. 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
Two light bulbs produce light (useful) and heat (waste).
- Incandescent bulb: Pin=100 W electrical, Pout=5 W light.
- LED bulb: Pin=20 W electrical, Pout=8 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 7
A heat engine takes in Ein=5000 J of thermal energy from fuel and produces Eout=1400 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 8
A solar panel receives Ein=2000 J of solar radiation in a short time interval and produces Eout=320 J of electrical energy. The rest becomes thermal energy in the panel. How much energy is lost as waste heat, Elost?
Use Ein=Eout+Elost.
- 1680 J (correct answer)
- 2320 J
- 320 J
- 620 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 9
A battery charging system stores Eout=760 J of chemical energy while Ein=800 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 10
An electric motor takes in electrical power Pin=500W and delivers useful mechanical power Pout=400W. The rest is lost as thermal energy in the motor. What is the motor's efficiency (as a percentage)?
Use η=PinPout×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=total energy inputuseful energy output×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 (Einput=Euseful+Ewaste), but not all input energy converts to the desired useful form. For this device, the input power is Pin=500 W and the useful output power is Pout=400 W, so efficiency η=500400×100%=80%. The remaining power Pwaste=500−400=100 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 (output/input)×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=(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
Two light bulbs convert electrical power into light (useful) and heat (waste). Bulb 1 has Pin=100 W and produces Plight=5 W. Bulb 2 has Pin=20 W and produces Plight=8 W. Which bulb is more efficient at producing light, and by how many percentage points?
Use: η=(PinPlight)×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 12
A heat engine takes in thermal energy Ein=5000 J from fuel and produces useful mechanical work Eout=1500 J. What is the engine efficiency η (as a percentage)?
Use: η=(EinEout)×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 13
An incandescent light bulb uses Pin=100W of electrical power and produces Pout=5W of useful light. The rest is released as heat. How much power is lost as thermal energy, Plost?
- 105W
- 95W (correct answer)
- 20W
- 5W
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 14
A battery charger transfers Ein=1500J of electrical energy to a rechargeable battery. After charging, the battery has stored Eout=1200J of chemical energy. The rest is lost mainly as heat. How much energy is lost, Elost?
- 300J (correct answer)
- 2700J
- 1200J
- 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 15
A device converts energy with Ein=3600J and produces useful output Eout=900J. 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 16
A battery charging system receives Ein=4000 J of electrical energy. Measurements show Elost=600 J is dissipated as heat. How much useful chemical energy is stored in the battery, Eout?
Use: Ein=Eout+Elost
- 4600 J
- 3400 J (correct answer)
- 2400 J
- 0.85 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 17
A solar panel receives Ein=2000 J of solar radiation in a short time interval and produces Eout=320 J of useful electrical energy. How much energy is lost as waste (mostly thermal), Elost?
Use: Ein=Eout+Elost
- 1680 J (correct answer)
- 2320 J
- 640 J
- 620 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 18
A solar panel produces Pout=90 W of electrical power from Pin=600 W of incoming solar power. What percentage of the input power is lost as waste (mostly thermal)?
Use: Plost=Pin−Pout and % lost=(PinPlost)×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 19
A solar panel operates at an efficiency of 18%. If it receives Ein=3000 J of solar radiation, how much useful electrical energy Eout does it produce?
Use: Eout=ηEin with η as a decimal.
- 540 J (correct answer)
- 166.7 J
- 3000 J
- 5400 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 20
A heat engine produces Pout=300 W of useful mechanical power at an efficiency of 30%. What input thermal power Pin is required?
Use: η=(PinPout)×100%.
- 100 W
- 900 W
- 1000 W (correct answer)
- 0.90 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).