Evaluate Device Energy Efficiency - Grade 11 Physics

1. What is the overall wall-to-mechanical efficiency, and how much wall energy is required to produce $252\ \mathrm{kJ}$ of mechanical work?
   Answer: $42\%$ and $600\ \mathrm{kJ}$ When a system has multiple stages of energy conversion, the overall efficiency is simply the product of all individual efficiencies — each stage only works with what the previous stage passed along. Here, you have three stages: charging (80%), battery-to-motor delivery (75%), and motor conversion (70%). Multiply them together: $\eta_{\text{overall}} = 0.80 \times 0.75 \times 0.70 = 0.42 = 42\%$ This means only 42 cents of every dollar of wall energy becomes useful mechanical work. To find the required wall energy, rearrange the efficiency formula $\eta = \frac{W_{\text{out}}}{E_{\text{in}}}$: $E_{\text{in}} = \frac{W_{\text{out}}}{\eta} = \frac{252\ \mathrm{kJ}}{0.42} = 600\ \mathrm{kJ}$ That confirms answer A — 42% efficiency and 600 kJ of wall energy required. Choice B (52.5%) likely comes from averaging the three efficiencies instead of multiplying them — a classic trap. Averaging gives $\frac{0.80+0.75+0.70}{3} = 0.75$... actually it produces 75%, so B may result from only multiplying two of the three stages. Choice C (60%) comes from multiplying just two stages, perhaps 0.80 × 0.75, ignoring the motor step entirely. Choice D gets the efficiency right (42%) but then multiplies instead of divides: $252 \times 0.42 \approx 105.8\ \mathrm{kJ}$, which is backwards logic. Study tip: In multi-stage efficiency problems, always multiply efficiencies (never add or average), and remember that wall energy must be greater than the useful output — so always divide output by efficiency, not the other way around.
2. What input power should the LED use, and how much electrical energy will the replacement save in one year?
   Answer: $15\ \mathrm{W}$ and $65.7\ \mathrm{kWh}$ When a question involves energy conversion efficiency, your key relationship is: useful output = efficiency × total input. Here, the incandescent produces $60 \times 0.10 = 6\ \mathrm{W}$ of useful visible light. The LED must match this output at 40% efficiency, so its required input power is $P = \frac{6\ \mathrm{W}}{0.40} = 15\ \mathrm{W}$. That confirms the first part of answer A. For the annual energy saving, calculate the energy each lamp consumes over one year. The incandescent uses $60\ \mathrm{W} \times 4.0\ \mathrm{h/day} \times 365\ \mathrm{days} = 87{,}600\ \mathrm{Wh} = 87.6\ \mathrm{kWh}$. The LED uses $15\ \mathrm{W} \times 4.0 \times 365 = 21{,}900\ \mathrm{Wh} = 21.9\ \mathrm{kWh}$. The saving is $87.6 - 21.9 = 65.7\ \mathrm{kWh}$, matching answer A exactly. Looking at the distractors: B (24 W) likely comes from mistakenly dividing 6 W by a wrong efficiency factor, perhaps confusing the two percentages. C (6 W) is a classic trap — it uses the useful light output directly as the LED's input power, forgetting to account for the LED's own conversion losses. D (40 W) ignores efficiency entirely and simply assumes the LED draws power equal to its efficiency percentage times the original lamp, which is a fundamental misapplication of the concept. As a study tip, always track two separate quantities — useful output and total input — and never confuse them. Efficiency questions almost always trap students who skip the step of converting useful output back into required input.
3. What is the loudspeaker's efficiency, and how much sound energy is produced during the test?
   Answer: $7.5\%$ and $3.60\ \mathrm{kJ}$ When a system has multiple components connected in series, the overall efficiency is the product of each component's individual efficiency — not just one of them. Keep this in mind whenever a problem breaks a system into stages. Here, the amplifier transfers $80\%$ of wall-outlet energy to the loudspeaker. The complete system (amplifier + loudspeaker together) converts $6.0\%$ of wall-outlet energy into sound. So the loudspeaker's efficiency describes how well it converts the energy it receives into sound. Using the relationship: $\eta_{\text{system}} = \eta_{\text{amp}} \times \eta_{\text{speaker}}$ $0.06 = 0.80 \times \eta_{\text{speaker}}$ $\eta_{\text{speaker}} = \frac{0.06}{0.80} = 0.075 = 7.5\%$ For the sound energy produced, first find the total energy drawn from the outlet over $5.0\ \mathrm{min} = 300\ \mathrm{s}$: $E_{\text{in}} = P \times t = 200\ \mathrm{W} \times 300\ \mathrm{s} = 60{,}000\ \mathrm{J}$ Then apply the system's overall efficiency: $E_{\text{sound}} = 0.06 \times 60{,}000 = 3{,}600\ \mathrm{J} = 3.60\ \mathrm{kJ}$ This confirms C is correct. Choice A incorrectly subtracts efficiencies ($0.80 - 0.06 \approx 0.048$) rather than dividing, and produces a wrong energy value. Choice B uses $6.0\%$ as the loudspeaker's efficiency directly — confusing the system efficiency with the speaker's individual efficiency. Choice D divides backwards ($0.06 \div 0.80$) to get the loudspeaker efficiency and then misapplies it to the energy calculation. As a study tip: always identify which stage's input energy you're calculating efficiency against. Draw a flow diagram — wall outlet → amplifier → loudspeaker → sound — to keep the stages and their inputs/outputs organized.
4. Which evaluation is best supported by the measurements?
   Answer: The efficiency is $87.5\%$, and about $63\ \mathrm{kJ}$ is transferred elsewhere. When a question asks you to evaluate a device's efficiency, your job is to compare the useful energy output to the total energy input — and check whether any energy is "lost" to the surroundings. Start by calculating the useful energy absorbed by the water: $Q_{useful} = mc\Delta T = 1.5 \times 4200 \times (90 - 20) = 441{,}000\ \mathrm{J} = 441\ \mathrm{kJ}$ Next, convert the electrical input into joules: $Q_{input} = 0.140\ \mathrm{kWh} \times 3{,}600{,}000\ \mathrm{J/kWh} = 504{,}000\ \mathrm{J} = 504\ \mathrm{kJ}$ Now calculate efficiency: $\eta = \frac{Q_{useful}}{Q_{input}} \times 100\% = \frac{441}{504} \times 100\% \approx 87.5\%$ The energy transferred elsewhere (wasted) is $504 - 441 = 63\ \mathrm{kJ}$. This confirms D — the kettle falls just short of the manufacturer's 90% claim, with about 63 kJ lost to the environment (heat escaping through the kettle's body, steam, etc.). Choice A is physically impossible — efficiency can never exceed 100%, as that would violate energy conservation. Choice B gets the wasted energy right (63 kJ ≈ 50 kJ is a rough match, but not really), but the 90.0% figure is fabricated — it matches the manufacturer's claim, not the measured data. Choice C uses the correct waste figure but inverts the efficiency fraction, dividing input by output instead of output by input. A useful habit: always compute both the efficiency and the wasted energy — questions often package both in the same answer choice to test whether you've done the full calculation.
5. What is the motor's efficiency during this lift?
   Answer: $79.1\%$, because both potential and kinetic energy gains are useful Whenever you see an efficiency question involving a motor doing mechanical work, remember: efficiency compares all useful energy output to total energy input. The trap is deciding what counts as "useful." Here, the motor does two things for the crate — it raises it (gaining gravitational PE) and accelerates it (gaining KE). Both represent real mechanical energy delivered to the crate, so both count as useful output. Calculating total input energy: $W_{input} = P \times t = 800\ \mathrm{W} \times 4.0\ \mathrm{s} = 3200\ \mathrm{J}$ Calculating useful output energy: $\Delta PE = mgh = 40 \times 9.8 \times 6.0 = 2352\ \mathrm{J}$ $\Delta KE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(40)(3.0)^2 = 180\ \mathrm{J}$ $W_{useful} = 2352 + 180 = 2532\ \mathrm{J}$ Efficiency: $\eta = \frac{2532}{3200} \times 100\% \approx 79.1\%$ This confirms B is correct. A is wrong because it ignores the kinetic energy gain — the crate is moving faster at the end, which required real energy from the motor. Leaving out $\Delta KE$ understates useful output. C is wrong because there's no physical reason to "count KE twice" — that's an invented rule with no basis in energy conservation. D is wrong because it flips the efficiency formula upside down (dividing input by output instead of output by input), which would give a value greater than 100% — physically impossible for a real motor. Study tip: Always list every form of mechanical energy the object gains — PE, KE, or both — and sum them as your useful output before dividing by total electrical input.
6. A student states that the refrigerator is $300\%$ efficient. Which evaluation of this statement is most accurate?
   Answer: The value $300\%$ is a performance coefficient, not an ordinary energy-conversion efficiency. Whenever you see a question involving refrigerators or heat pumps, recognize that these are heat-transfer devices, not energy-conversion devices like motors or engines. This distinction completely changes how we measure their "efficiency." For a refrigerator, the relevant performance measure is called the Coefficient of Performance (COP), defined as the heat removed from the cold reservoir divided by the work input: $\text{COP} = \frac{Q_{\text{removed}}}{W_{\text{input}}} = \frac{600\ \text{kJ}}{200\ \text{kJ}} = 3$ Expressed as a percentage-style ratio, this gives 300%. This is not an ordinary efficiency — it's a dimensionless performance ratio that can legally exceed 1 (or 100%) because the device moves thermal energy rather than converting it. Energy is still conserved: $600 + 200 = 800\ \text{kJ}$, perfectly consistent with the first law. This confirms C is correct. A is wrong because ordinary energy-conversion efficiency is always capped at 100% — you cannot get more useful energy out than you put in. Only heat-pump-style devices exceed this apparent limit, and even then, it's a COP, not a true efficiency. B is wrong because no energy is created. The 800 kJ released equals the 600 kJ absorbed plus the 200 kJ of electrical work input — energy is conserved, not manufactured. D is wrong because comparing output heat to input heat ($\frac{600}{800} = 75\%$) has no standard physical meaning for a refrigerator and doesn't correspond to any recognized performance metric. Study tip: Always distinguish between COP (for refrigerators/heat pumps) and efficiency (for engines/converters) — exam questions frequently use the word "efficiency" loosely to test whether you know the difference.
7. Which set correctly gives the panel efficiency, pump efficiency, and overall sunlight-to-water efficiency, respectively?
   Answer: $18\%$, $60\%$, and $10.8\%$ Whenever you see a multi-stage energy conversion problem, your job is to calculate efficiency at each stage separately, then multiply (or recalculate) for the overall system. Efficiency is always $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%$. Start with the solar panel: it receives $1500\ \mathrm{W}$ and outputs $270\ \mathrm{W}$ of electrical power, so its efficiency is $\frac{270}{1500} \times 100\% = 18\%$. Next, the pump: it receives $270\ \mathrm{W}$ electrically and delivers $162\ \mathrm{W}$ as gravitational potential energy, giving $\frac{162}{270} \times 100\% = 60\%$. Finally, the overall sunlight-to-water efficiency uses the original input and the final useful output: $\frac{162}{1500} \times 100\% = 10.8\%$. This confirms answer A: $18\%$, $60\%$, and $10.8\%$. Answer B swaps the panel and pump efficiencies — a classic label-mixing trap where you get the right numbers but assign them to the wrong device. Answer C gets the panel and pump efficiencies right but calculates the overall efficiency incorrectly as $18\% + 60\% = 78\%$; you cannot simply add stage efficiencies together. Answer D confuses the overall efficiency ($10.8\%$) with the panel efficiency, placing the values in the wrong order entirely. A useful strategy: always identify which device's input and output you're working with before plugging in numbers. Label each calculation as you go — "panel in, panel out," "pump in, pump out" — to avoid mixing values across stages.
8. What is the efficiency of the regenerative braking system in converting the mechanical energy available to the brakes into stored electrical energy?
   Answer: $60\%$, subtracting the hill-climbing energy from the kinetic-energy decrease When tackling efficiency problems involving energy conversion, always ask: what energy was actually available to the system doing the converting? Here, the regenerative braking system only captures energy that flows through the brakes — it cannot capture energy the car already spent climbing the hill. Start by identifying all energy changes. The car's kinetic energy decreases by $\Delta KE = \frac{1}{2}(1000)(20^2 - 10^2) = 150{,}000\ \mathrm{J} = 150\ \mathrm{kJ}$. However, $40\ \mathrm{kJ}$ of that went directly into gravitational potential energy: $\Delta PE = mgh = (1000)(10)(4.0) = 40\ \mathrm{kJ}$. The energy actually available to the brakes is $150 - 40 = 110\ \mathrm{kJ}$. The battery stores $66\ \mathrm{kJ}$, giving efficiency $= \frac{66}{110} \approx 60\%$. That confirms C is correct. Choice A uses the full initial kinetic energy ($\frac{1}{2}mv_0^2 = 200\ \mathrm{kJ}$) as the denominator, which ignores both the final kinetic energy remaining and the hill-climbing energy — a double error in identifying the input. Choice B uses the total $\Delta KE = 150\ \mathrm{kJ}$ as the denominator, which is closer but still wrong: it treats the $40\ \mathrm{kJ}$ used for climbing as if it were available to the brakes, inflating the denominator and deflating the efficiency. Choice D adds the hill-climbing energy to the battery gain, which makes no physical sense — the gravitational PE was stored in the car's position, not in the battery. The key strategy: efficiency always compares output to the relevant input. Subtract any energy diverted to other purposes before calculating what the converter actually had to work with.
9. Which evaluation correctly compares their overall daily efficiencies, including standby energy?
   Answer: Machine B is better, with about $72.3\%$ compared with Machine A's $62.5\%$. Whenever you see an efficiency question that includes standby power, remember that overall efficiency must account for all energy consumed, not just during active operation. Here's how to calculate each machine's true daily efficiency: $\eta = \frac{E_{\text{useful}}}{E_{\text{total input}}}$, where $E_{\text{total}}$ includes both active and standby energy. Machine A draws $500\ \text{W} \times 2.0\ \text{h} = 1000\ \text{Wh}$ while active, plus $20\ \text{W} \times 22\ \text{h} = 440\ \text{Wh}$ in standby, giving a total of $1440\ \text{Wh}$. With $900\ \text{Wh}$ of useful output: $\eta_A = \frac{900}{1440} \approx 62.5\%$. Machine B must draw $\frac{900\ \text{Wh}}{0.75} = 1200\ \text{Wh}$ during its active phase, plus $2.0\ \text{W} \times 22\ \text{h} = 44\ \text{Wh}$ in standby, for a total of $1244\ \text{Wh}$. Thus $\eta_B = \frac{900}{1244} \approx 72.3\%$. Machine B wins — confirming answer B. Answer A is wrong because Machine A's 90% applies only during active operation; adding standby consumption drags the true efficiency down to ~62.5%. Answer C is a trap: equal useful output does not mean equal efficiency — you must compare output to total input. Answer D gets Machine A's efficiency right but incorrectly calculates Machine B's, underestimating it as 60%. Study tip: On efficiency problems, always flag any idle or standby power mentioned in the passage — it's almost always there specifically to lower the apparent efficiency and test whether you include it in your denominator.
10. What is the device's efficiency over the complete operating interval?
    Answer: $40\%$, weighting each output power by its operating time Whenever you see an efficiency problem where conditions change over time, your instinct should be to work with total energy, not instantaneous power values alone. Efficiency is always the ratio of useful output energy to total input energy over the same interval. Here's how to work through it correctly. The total input energy is constant throughout: $P_{\text{in}} \times t = 500 \times 120 = 60{,}000\ \text{J}$. The useful output energy must be calculated in two separate segments. During the first 30 s: $350 \times 30 = 10{,}500\ \text{J}$. During the remaining 90 s: $150 \times 90 = 13{,}500\ \text{J}$. Total useful output: $10{,}500 + 13{,}500 = 24{,}000\ \text{J}$. Efficiency is therefore $\dfrac{24{,}000}{60{,}000} = 0.40 = 40\%$, confirming answer B. Choice A uses only the lower output power (150 W) as if it applied for the full 120 s, giving $\frac{150}{500} = 30\%$. This ignores the higher-output phase entirely. Choice C averages the two power values equally — $\frac{(350+150)/2}{500} = 50\%$ — which would only be valid if both powers lasted the same duration; here they don't. Choice D uses only the higher output power (350 W) for the full interval, giving 70%, which ignores the longer, lower-output phase. The key study tip: never average rates without accounting for time. When durations differ, convert everything to total energy first, then take the ratio. This principle appears in efficiency, average speed, and weighted-average problems across all of physics.

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