Evaluate Device Energy Efficiency - Physics

Two light bulbs convert electrical power into light (useful) and heat (waste). - Incandescent bulb: $P_{in}=100\ \text{W}$ , useful light output $P_{out}=5\ \text{W}$ . - LED bulb: $P_{in}=20\ \text{W}$ , useful light output $P_{out}=8\ \text{W}$ . Which bulb is more efficient, and by how many percentage points is it more efficient? (Report the difference in efficiency in percentage points.)
Answer: LED by $35$ percentage points 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, η = (5/100) × 100% = 5%; for the LED, η = (8/20) × 100% = 40%, so the LED is more efficient by 35 percentage points. The waste is higher in incandescent due to greater thermal losses from resistance in the filament compared to LEDs. Choice C is correct because it accurately compares efficiencies showing which device converts larger fraction. Choice B makes an arithmetic error by reversing the comparison. 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.).
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, $E_{lost}$ ?
Answer: $300\,\text{J}$ 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.).
A heat engine takes in $E_{in}=8000\ \text{J}$ of thermal energy from fuel and produces $E_{out}=2400\ \text{J}$ of useful mechanical work. The remainder leaves as waste heat. What fraction of the input energy is wasted as heat, expressed as a percentage? (Use $E_{lost}=E_{in}-E_{out}$ , then $\%\text{wasted}=(E_{lost}/E_{in})\times 100\%$ .)
Answer: $70\%$ 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 = 8000 J and the useful output energy is E_out = 2400 J, so the lost energy E_lost = E_in - E_out = 8000 - 2400 = 5600 J, and the percentage wasted = (E_lost/E_in) × 100% = (5600/8000) × 100% = 70%. The remaining energy E_waste = 5600 J is lost as waste heat, which is characteristic of heat engines where incomplete combustion and thermal losses limit efficiency. Choice B is correct because it properly calculates the waste percentage as (E_lost/E_in) × 100% = (5600/8000) × 100% = 70%. Choice A calculates the efficiency (2400/8000 = 30%) instead of waste percentage; Choice C makes an arithmetic error; Choice D 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. Higher efficiency means more input converts to useful output, less wasted.
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 $E_{in}=E_{out}+E_{lost}$ .
Answer: $1680\ \text{J}$ 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.).
A battery charger takes in $E_{in}=900\ \text{J}$ of electrical energy and stores $E_{out}=810\ \text{J}$ as useful chemical energy in the battery. The rest is lost as heat. What is the charging efficiency $\eta$ (in %)?
Answer: $90\%$ 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 = 900 J and the useful output energy is E_out = 810 J, so efficiency η = (E_out/E_in) × 100% = (810/900) × 100% = 90%. The remaining energy E_waste = E_in - E_out = 90 J is lost as heat, which is characteristic of battery chargers where electrical resistance causes losses. Choice A 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.).
A battery-charging system takes in $P_{in}=250\ \text{W}$ of electrical power. If $P_{lost}=50\ \text{W}$ is dissipated as heat, what is the useful power stored as chemical energy, $P_{out}$ ? Use: $P_{in}=P_{out}+P_{lost}$ .
Answer: $200\ \text{W}$ 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 power is P_in = 250 W and the power lost as heat is P_lost = 50 W, so using energy conservation P_in = P_out + P_lost, we get P_out = P_in - P_lost = 250 - 50 = 200 W. The remaining power P_out = 200 W is stored as chemical energy in the battery, which is characteristic of charging systems where electrical resistance and electrochemical inefficiencies generate heat. Choice B is correct because it properly calculates useful output power as P_out = P_in - P_lost = 250 - 50 = 200 W. Choice A (300 W) incorrectly adds instead of subtracting (250 + 50); Choice C (50 W) confuses waste power with useful output; Choice D (0.80 W) appears to calculate efficiency as decimal instead of power. 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. Energy conservation requires input = output + waste for any device.
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?
Answer: $5\%$ 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.).
A battery charger takes in $E_{in}=5000\ \text{J}$ of electrical energy. It stores $E_{out}=4500\ \text{J}$ as useful chemical energy in the battery, with the rest lost as heat. What is the charging efficiency $\eta$ (in %)?
Answer: $90\%$ 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 = 5000 J and the useful output energy is E_out = 4500 J, so efficiency η = (E_out/E_in) × 100% = (4500/5000) × 100% = 90%. The remaining energy E_waste = E_in - E_out = 500 J 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% = (4500/5000) × 100% = 90%. Choice B incorrectly inverts the ratio calculating input/output instead of output/input, giving 110% which violates conservation; Choice C calculates the waste percentage (500/5000 = 10%) instead of efficiency; Choice D 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. Higher efficiency means more input converts to useful output, less wasted.
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. What is the motor's efficiency $\eta$ as a percentage? (Use $\eta=(P_{out}/P_{in})\times 100\%$ .)
Answer: $80\%$ 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, the input power is P_in = 500 W and the useful output power is P_out = 400 W, so efficiency η = (P_out/P_in) × 100% = (400/500) × 100% = 80%. The remaining power P_waste = P_in - P_out = 100 W is lost as thermal energy, which is characteristic of electric motors where electrical resistance in windings and mechanical friction cause heating. Choice C is correct because it properly calculates efficiency as (output/input) × 100% = (400/500) × 100% = 80%. Choice A incorrectly inverts the ratio calculating input/output instead of output/input, giving 125% which violates conservation; Choice B makes an arithmetic error yielding 0.80%; Choice D calculates the waste percentage (100/500 = 20%) 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.
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 %)?
Answer: $16\%$ 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.
A heat engine produces useful work at a rate of $P_{out}=300\ \text{W}$ while taking in thermal power $P_{in}=1200\ \text{W}$ . The remainder is rejected as waste heat. What is the waste power $P_{lost}$ ?
Answer: $900\ \text{W}$ 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 = 1200 W and the useful output power is P_out = 300 W, so efficiency η = (300/1200) × 100% = 25%. The remaining power P_waste = 1200 - 300 = 900 W is lost as thermal energy, which is characteristic of heat engines where incomplete combustion and thermodynamic limits cause losses. Choice A is correct because it correctly identifies waste power as P_in - P_out. Choice B incorrectly adds powers 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.).
A heat engine receives $E_{in} = 5000\ \text{J}$ of thermal energy from burning fuel and produces $E_{out} = 1500\ \text{J}$ of useful mechanical work. The rest is expelled as waste heat. What percentage of the input energy is wasted as thermal energy? Use: $E_{lost}=E_{in}-E_{out}$ and $\%\text{ wasted} = \left(\frac{E_{lost}}{E_{in}}\right)\times 100\%$ .
Answer: $70\%$ 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 the waste energy E_lost = E_in - E_out = 5000 - 1500 = 3500 J, and the percentage wasted = (E_lost/E_in) × 100% = (3500/5000) × 100% = 0.70 × 100% = 70%. The remaining energy E_waste = 3500 J is expelled as waste heat, which is characteristic of heat engines where incomplete combustion and thermodynamic limits prevent full conversion to mechanical work. Choice B is correct because it properly calculates waste percentage as (E_lost/E_in) × 100% = (3500/5000) × 100% = 70%. Choice A (30%) incorrectly calculates the efficiency (1500/5000) instead of waste percentage; Choice C (130%) makes arithmetic error giving >100%; Choice D (0.70%) 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. Heat engines typically waste 60-70% of input energy due to fundamental thermodynamic limits.
A light bulb receives $E_{in}=1200\ \text{J}$ of electrical energy in a brief period and produces $E_{out}=60\ \text{J}$ of useful light energy (the rest becomes thermal energy). What fraction of the input energy is wasted as heat? Give your answer as a percentage.
Answer: $95\%$ 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 = 1200 J and the useful output energy is E_out = 60 J, so efficiency η = (E_out/E_in) × 100% = (60/1200) × 100% = 5%. The remaining energy E_waste = E_in - E_out = 1140 J is lost as thermal energy, which is characteristic of light bulbs where electrical resistance in the filament causes heat losses, and percent waste = (1140/1200) × 100% = 95%. Choice B is correct because it correctly computes the waste percentage as ((input - output)/input) × 100%. Choice C inverts the ratio and makes an arithmetic error, 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.).
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 (mostly as thermal energy). What is the motor's efficiency $\eta$ as a percentage? Use: $\eta = \left(\frac{P_{out}}{P_{in}}\right)\times 100\%$
Answer: $80\%$ 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 electric motor, the input power is P_in = 500 W and the useful output power is P_out = 400 W, so efficiency η = (P_out/P_in) × 100% = (400/500) × 100% = 0.80 × 100% = 80%. The remaining power P_waste = P_in - P_out = 500 - 400 = 100 W is lost as thermal energy, which is characteristic of electric motors where electrical resistance and friction in bearings cause heating. Choice C is correct because it properly calculates efficiency as (400/500) × 100% = 80%. Choice A (125%) inverts the ratio calculating input/output instead of output/input, giving >100% which violates conservation; Choice B (0.80%) forgets to multiply by 100% to convert decimal to percentage; Choice D (20%) calculates the waste percentage instead of efficiency. To calculate efficiency: (1) identify useful output power, (2) identify total input power, (3) compute η = (output/input) × 100%, (4) verify η between 0-100%, (5) waste power = input - output. Higher efficiency means more input converts to useful output, less wasted.
An electric motor converts electrical energy into mechanical work. The motor takes in $P_{in}=500\ \text{W}$ of electrical power and delivers $P_{out}=400\ \text{W}$ of useful mechanical power (the rest becomes thermal energy in the motor). What is the motor's efficiency $\eta$ as a percentage? Use: $\eta=\dfrac{P_{out}}{P_{in}}\times 100\%$ .
Answer: $80\%$ 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 electric motor, the input power is P_in = 500 W and the useful output power is P_out = 400 W, so efficiency η = (P_out/P_in) × 100% = (400/500) × 100% = 80%. The remaining power P_waste = P_in - P_out = 100 W is lost as thermal energy, which is characteristic of electric motors where electrical resistance and friction cause losses. Choice B 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.).
A solar panel produces $P_{out}=150\ \text{W}$ of useful electrical power when sunlight provides $P_{in}=1000\ \text{W}$ of radiant power to the panel. Assuming the rest becomes thermal energy, what is the panel efficiency $\eta$ (in %)?
Answer: $15\%$ 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 power is P_in = 1000 W and the useful output power is P_out = 150 W, so efficiency η = (P_out/P_in) × 100% = (150/1000) × 100% = 15%. The remaining power P_waste = P_in - P_out = 850 W is lost as thermal energy, which is characteristic of solar panels where incomplete conversion and reflection cause losses. Choice A 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.).
A heat engine takes in $E_{in}=6000\ \text{J}$ of thermal energy from burning fuel and produces $E_{out}=1800\ \text{J}$ of useful mechanical work. The remaining energy is rejected as waste heat. What percentage of the input energy becomes waste heat? Use: $E_{lost}=E_{in}-E_{out}$ and percent waste $=\dfrac{E_{lost}}{E_{in}}\times 100\%$ .
Answer: $70\%$ 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 = 6000 J and the useful output energy is E_out = 1800 J, so efficiency η = (E_out/E_in) × 100% = (1800/6000) × 100% = 30%. The remaining energy E_waste = E_in - E_out = 4200 J is lost as waste heat, which is characteristic of heat engines where incomplete combustion and thermal losses occur, and percent waste = (4200/6000) × 100% = 70%. Choice B is correct because it correctly identifies waste energy as E_in - E_out and computes the percentage. Choice C inverts the ratio and makes an arithmetic error, 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.).
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: $\eta = \left(\frac{P_{out}}{P_{in}}\right)\times 100\%$ .
Answer: $1000\ \text{W}$ 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).
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.
Answer: $540\ \text{J}$ 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).
Two bulbs are tested: - Incandescent: $P_{in}=100\,\text{W}$ , useful light $P_{out}=5\,\text{W}$ - LED: $P_{in}=20\,\text{W}$ , useful light $P_{out}=8\,\text{W}$ Which bulb is more efficient, and by how many percentage points (LED efficiency minus incandescent efficiency)?
Answer: LED by $35$ percentage points 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 these devices, the incandescent bulb has η = (5/100) × 100% = 5% and the LED has η = (8/20) × 100% = 40%, so the difference is 40% - 5% = 35 percentage points, with LED being more efficient. The waste in incandescent is higher due to greater thermal losses from resistance in the filament compared to LED's semiconductor efficiency. Choice B is correct because it accurately compares efficiencies showing which device converts larger fraction. Choice C is wrong because it makes arithmetic error in difference 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. Higher efficiency means more input converts to useful output, less wasted. No device achieves 100% due to unavoidable losses (friction, resistance, etc.).

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