Model Energy Transfer Between Objects - Grade 11 Physics

1. Which statement correctly models the energy transfer before thermal equilibrium is reached?
   Answer: About $9.45\times10^3\ \mathrm{J}$ transfers from aluminum to water, producing a final temperature of $27.5\ ^\circ\mathrm{C}$ When two objects at different temperatures are placed in contact inside an insulated container, heat flows from the hotter object to the cooler one until they reach the same temperature — called thermal equilibrium. The governing principle is conservation of energy: heat lost by the hot object equals heat gained by the cool one. Use $Q = mc\Delta T$ for each substance, then set $Q_{\text{lost}} = Q_{\text{gained}}$ to solve for the final temperature $T_f$. Setting up the equation: $m_{\text{Al}} c_{\text{Al}} (80.0 - T_f) = m_{\text{w}} c_{\text{w}} (T_f - 20.0)$, which gives $(0.200)(900)(80.0 - T_f) = (0.300)(4200)(T_f - 20.0)$. Simplifying: $180(80.0 - T_f) = 1260(T_f - 20.0)$, so $14400 - 180T_f = 1260T_f - 25200$, yielding $T_f = \frac{39600}{1440} \approx 27.5\ ^\circ\mathrm{C}$. The energy transferred is $Q = (0.200)(900)(80.0 - 27.5) \approx 9450\ \mathrm{J} = 9.45 \times 10^3\ \mathrm{J}$, flowing from aluminum to water. That confirms A is correct. B gets the final temperature right but reverses the direction of heat flow — heat always travels from hot to cold, never the other way. C uses $50.0\ ^\circ\mathrm{C}$ (the arithmetic average of 80 and 20), which only works when $mc$ is identical for both substances — it isn't here. D assumes no temperature change in water, which violates energy conservation entirely. A helpful tip: always check both the magnitude and the direction of energy transfer. A correct number with the wrong direction is still a wrong answer.
2. Which statement completes the energy-transfer model for the crate and ramp?
   Answer: $200\ \mathrm{J}$ is transferred to thermal energy by friction between the crate and ramp Whenever you see a question about work and energy on a ramp with friction, your first instinct should be to apply conservation of energy. The total energy input must equal the total energy output — it can't disappear, it just changes form. Here, the worker does $900\ \mathrm{J}$ of work on the crate. That energy has to go somewhere. You can account for it by adding up every form of energy the crate gains: $W_{input} = \Delta PE + \Delta KE + \Delta E_{thermal}$ $900\ \mathrm{J} = 600\ \mathrm{J} + 100\ \mathrm{J} + \Delta E_{thermal}$ $\Delta E_{thermal} = 900 - 600 - 100 = 200\ \mathrm{J}$ This confirms C — friction between the crate and ramp converts $200\ \mathrm{J}$ into thermal energy, perfectly balancing the energy ledger. A is wrong because it uses $800\ \mathrm{J}$, which ignores both the gravitational PE gain and the KE gain. The reasoning ("kinetic energy is temporary") is also a misconception — kinetic energy is a real, accountable form of energy, not something to discard. B is wrong because $300\ \mathrm{J}$ comes from subtracting only the $600\ \mathrm{J}$ PE gain, ignoring the $100\ \mathrm{J}$ increase in kinetic energy. All energy gains must be accounted for, not just the ones deemed "useful." D is wrong because $100\ \mathrm{J}$ is the kinetic energy gained, not the thermal energy produced — these are two separate outputs. Study tip: Always write the full energy equation first — input equals the sum of all outputs. This prevents you from accidentally leaving out a term.
3. Which statement best models the missing energy when the system is defined as the ball and Earth only?
   Answer: $30\ \mathrm{J}$ leaves the system through work done on the surrounding air Whenever you see an energy question involving a falling object, your first move should be to apply conservation of energy — but pay close attention to how the system is defined. Here, the system is only the ball and Earth, which means the surrounding air is outside the system. Any energy transferred to that outside air counts as energy leaving the system. The numbers tell the story clearly: gravitational PE drops by $100\ \mathrm{J}$, but KE only rises by $70\ \mathrm{J}$. That leaves $30\ \mathrm{J}$ unaccounted for. Since air resistance acts on the ball, the ball does work on the surrounding air molecules — transferring $30\ \mathrm{J}$ out of the ball-Earth system entirely. This makes D correct: the missing $30\ \mathrm{J}$ exits the system through work done on the surrounding air (drag). A is wrong because energy is never "destroyed" — this violates conservation of energy entirely. The $30\ \mathrm{J}$ isn't gone; it's transferred elsewhere. B is wrong because gravitational PE is already accounted for in the $100\ \mathrm{J}$ decrease — it can't simultaneously remain stored as additional gravitational energy. That contradicts the given information. C flips the numbers entirely, claiming $70\ \mathrm{J}$ leaves the system. But we know $70\ \mathrm{J}$ stayed inside the system as kinetic energy — only $30\ \mathrm{J}$ was transferred out. Study tip: Always identify the system boundary first. Energy that crosses that boundary is transferred, not destroyed — and the direction of the gap (PE lost vs. KE gained) tells you exactly how much left the system.
4. Which model best accounts for the energy transferred during the lift?
   Answer: $980\ \mathrm{J}$ becomes gravitational energy, and $420\ \mathrm{J}$ becomes thermal energy When a motor lifts a load that starts and ends at rest, you're dealing with energy conservation: all the electrical energy input must be accounted for as either useful gravitational potential energy or wasted thermal energy (from friction, heat in motor windings, etc.). No kinetic energy remains since the load is stationary at both ends. Start by calculating the gravitational potential energy gained by the load: $E_{grav} = mgh = (20)(9.8)(5.0) = 980\ \mathrm{J}$ The motor received $1400\ \mathrm{J}$ total, so the energy that became thermal must be: $E_{thermal} = 1400 - 980 = 420\ \mathrm{J}$ This matches A perfectly: $980\ \mathrm{J}$ goes to gravitational potential energy, and the remaining $420\ \mathrm{J}$ is lost as thermal energy due to inefficiency in the motor and mechanical system. B is wrong because it assumes the motor is 100% efficient — all $1400\ \mathrm{J}$ cannot become gravitational energy when the load only gains $980\ \mathrm{J}$ worth. Real motors always waste some energy. C flips the two values, assigning $420\ \mathrm{J}$ to gravity and $980\ \mathrm{J}$ to heat. This contradicts the calculated $mgh$, which is unambiguously $980\ \mathrm{J}$. D introduces kinetic energy as a final destination, but since the load ends at rest, there is zero net kinetic energy — this answer ignores the problem's explicit condition. A useful habit: always calculate $mgh$ first, then subtract from total input energy to find thermal losses. If kinetic energy is zero at start and finish, it simply doesn't appear in your final energy accounting.
5. How much energy is transferred to thermal energy in the pulley, rope, and surroundings during the motion?
   Answer: $8.7\ \mathrm{J}$, the net gravitational-energy decrease minus the total kinetic-energy increase Whenever energy is involved in a mechanics problem with moving objects, your go-to tool is the conservation of energy: the initial energy equals the final energy plus any losses to thermal energy (friction, air resistance, etc.). Here, the "lost" energy equals the net work done by gravity minus the kinetic energy actually gained. Start by calculating the net gravitational potential energy change. The 2.0 kg block descends 1.5 m, losing $\Delta PE_1 = (2.0)(9.8)(1.5) = 29.4\ \mathrm{J}$. The 1.0 kg block rises 1.5 m, gaining $\Delta PE_2 = (1.0)(9.8)(1.5) = 14.7\ \mathrm{J}$. The net gravitational energy released is $29.4 - 14.7 = 14.7\ \mathrm{J}$. Next, find the total kinetic energy gained: $KE = \frac{1}{2}(2.0)(2.0)^2 + \frac{1}{2}(1.0)(2.0)^2 = 4.0 + 2.0 = 6.0\ \mathrm{J}$. The thermal energy produced is therefore $14.7 - 6.0 = 8.7\ \mathrm{J}$, confirming answer B. Choice A confuses the final kinetic energy (6.0 J) with thermal losses — but kinetic energy is useful energy the system kept, not energy that was lost. Choice C mistakes the gross gravitational loss of the descending block alone (14.7 J) for thermal energy, ignoring that the rising block stored some energy back as potential energy. Choice D incorrectly adds kinetic energy to the net gravitational decrease, double-counting energy rather than subtracting. A useful strategy: thermal energy lost = (net energy input from gravity) − (energy stored as KE or PE). Always account for all energy transformations before labeling anything as "lost."
6. Which energy model is consistent with the data?
   Answer: The battery transfers $7.20\times10^3\ \mathrm{J}$, with $1.20\times10^3\ \mathrm{J}$ transferred elsewhere in the circuit Whenever you see a question combining voltage, current, and time, your first move should be calculating the total electrical energy supplied by the source using $E = P \cdot t = V \cdot I \cdot t$. Then apply energy conservation: every joule supplied must go somewhere. Here, the battery supplies $E = (12\ \mathrm{V})(2.0\ \mathrm{A})(5.0 \times 60\ \mathrm{s}) = (12)(2.0)(300) = 7.20 \times 10^3\ \mathrm{J}$. The device dissipates $6.00 \times 10^3\ \mathrm{J}$ as thermal energy, leaving $7.20 \times 10^3 - 6.00 \times 10^3 = 1.20 \times 10^3\ \mathrm{J}$ unaccounted for in the device alone. That remaining energy is transferred elsewhere in the circuit — for example, as heat in the battery's internal resistance or connecting wires. This matches D exactly, and it respects conservation of energy perfectly. A is wrong because it uses only $V \times I = 24\ \mathrm{W}$ and mistakes power (watts) for energy (joules) — forgetting to multiply by time entirely. B sets the supplied energy equal to the thermal output, which would only be true if the circuit were 100% efficient with no losses anywhere else — but conservation of energy doesn't allow the device to output exactly what the whole circuit receives unless nothing else dissipates energy. C invents energy out of nowhere: it claims the device creates thermal energy beyond what the battery supplies, violating conservation of energy entirely. Your go-to strategy: always calculate total electrical energy first with $E = VIt$, then use energy conservation as your check — energy in must equal energy out across the whole system.
7. Compared with pulse X, how much energy does pulse Y transfer past a fixed point on the rope?
   Answer: Twice as much, because power scales as amplitude squared, giving a factor of 4, which is then halved by the shorter duration When a wave pulse travels along a rope, the energy it carries depends on two things: the power it delivers and the duration over which it delivers that power. Power in a wave scales with the square of amplitude, so doubling the amplitude multiplies the power by $2^2 = 4$. Total energy is then: $E = P \times \Delta t$ For pulse Y: amplitude is doubled (power ×4) but duration is halved (×½), giving a total energy of $4 \times \frac{1}{2} = 2$ times that of pulse X. So pulse Y transfers twice as much energy — confirming that A is correct. Looking at why the other choices fail: B incorrectly treats the two factors as perfectly canceling. They don't — amplitude appears squared in the power expression, so you get a factor of 4 from amplitude, not 2. If amplitude only gave a factor of 2, then yes, it would cancel the ½ duration. That's the trap B sets. C ignores amplitude's contribution to energy entirely, claiming only duration matters — this is wrong because a higher-amplitude pulse delivers more power at every instant. D correctly identifies the amplitude-squared relationship but then dismisses duration entirely, forgetting that energy is power integrated over time; a pulse that ends sooner simply delivers energy for less time. A useful strategy: whenever a wave problem changes both amplitude and duration, handle them separately. First apply $P \propto A^2$, then multiply by the new duration. Students frequently forget the square on amplitude — always write out the factor explicitly before combining.
8. Assuming the melted water remains at $0\ ^\circ\mathrm{C}$, what mass of ice melts?
   Answer: $0.072\ \mathrm{kg}$, based on the energy actually transferred from the heater to the ice When a question involves a heater melting ice, your first instinct should be to carefully track which energy actually reaches the ice — not just the total electrical energy consumed. These are often different, and that distinction is the entire point of the question. Here, the heater runs at $100\ \mathrm{W}$ for $5.0\ \mathrm{min} = 300\ \mathrm{s}$, so the total electrical energy is $E_{total} = 100 \times 300 = 30{,}000\ \mathrm{J}$. However, only $80\%$ of that energy is actually transferred to the ice: $E_{ice} = 0.80 \times 30{,}000 = 24{,}000\ \mathrm{J}$. Since the water stays at $0\ ^\circ\mathrm{C}$, all this energy goes into the phase change. Using the latent heat equation $Q = mL_f$, you get $m = \frac{24{,}000}{3.34 \times 10^5} \approx 0.072\ \mathrm{kg}$. That confirms B is correct. Choice A is a classic efficiency trap — it uses the full $30{,}000\ \mathrm{J}$ without applying the $80\%$ factor, giving $0.090\ \mathrm{kg}$. The problem explicitly states that only $80\%$ transfers to the ice, so ignoring that step is a direct misread. Choice C compounds the error by treating the lost $20\%$ as extra melting energy — the opposite of what happens; lost energy never reaches the ice. Choice D misuses the heater's power value ($100\ \mathrm{W}$) as if it were energy in joules, skipping the time calculation entirely, which produces a nonsensical result. Your study tip: whenever a problem mentions an efficiency percentage, apply it immediately to find the useful energy before doing anything else. Writing $Q_{useful} = \eta \cdot P \cdot t$ as your first line will keep you from falling into the most common trap on these questions.
9. Which pair of energy transfers is predicted by this model?
   Answer: $108\ \mathrm{kJ}$ to the battery and $72\ \mathrm{kJ}$ to thermal energy When you see a question about regenerative braking, treat it as a kinetic energy partition problem: find the total energy change first, then split it using the given percentages. Start by calculating the decrease in kinetic energy: $\Delta KE = \frac{1}{2}mv_i^2 - \frac{1}{2}mv_f^2 = \frac{1}{2}(1200)(20^2) - \frac{1}{2}(1200)(10^2)$ $= 240{,}000 - 60{,}000 = 180{,}000\ \mathrm{J} = 180\ \mathrm{kJ}$ Now apply the 60/40 split: the battery receives $0.60 \times 180 = 108\ \mathrm{kJ}$, and the remaining $0.40 \times 180 = 72\ \mathrm{kJ}$ becomes thermal energy. That matches answer C. Choice A assigns the full $180\ \mathrm{kJ}$ to the battery and $120\ \mathrm{kJ}$ to thermal — these don't even add up to the correct total, so the student who picks this likely skipped computing $\Delta KE$ entirely and used the initial KE instead. Choice B reverses the percentages, giving 40% to the battery ($72\ \mathrm{kJ}$) and 60% to thermal ($108\ \mathrm{kJ}$) — a classic swap trap. Choice D uses values of $216\ \mathrm{kJ}$ and $144\ \mathrm{kJ}$, which sum to $360\ \mathrm{kJ}$; this comes from mistakenly adding the initial and final kinetic energies rather than subtracting them. Strategy tip: Always compute $\Delta KE$ as a subtraction before applying any efficiency percentage, and verify your two parts sum back to that total — it's a quick self-check that catches most errors on these problems.

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