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Thermo-physics · quiz worked solutions

Reference from the original Physics I textbook.

Tried the quiz first? Good. Here’s how to work through each answer. Follow the link beside a question to jump back to its original spot. Questions marked Extra exploration go beyond this AP Physics 1 course; Book controls questions explain the interface.

Heat Transfer · question 1

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Question: Lakes have a very rich nutrient reservoir at the bottom. This is due to organic matter accumulating on the bottom. In the spring, the lake “turnover” occurs where the water from the bottom rises to the top. What does this show?

Answer: Convection.

  1. Notice that water itself circulates through the lake. Why: Convection carries thermal energy and dissolved material with fluid motion.

  2. Choose convection rather than conduction or radiation. Why: The detailed turnover depends on density and temperature, but the clue specifically describes bulk transport.

Heat Transfer · question 2

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Question: Using area-normalized resistance R = L/k, what is R_iron/R_silver for a 1.0 m iron slab and a 2.0 m silver slab? Use k_iron = 67 and k_silver = 428 W/(m·K).

Answer: R_iron/R_silver = 214/67 ≈ 3.19.

  1. Use R = L/k to find R_iron = 1/67 and R_silver = 2/428. Why: These are area-normalized resistances, as specified.

  2. Divide (1/67)/(2/428) = 214/67. Why: A material’s conductivity alone is not enough when thicknesses differ.

Heat Transfer · question 3

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Question: Four layers I, II, III and IV have equal thickness and area and carry the same nonzero steady heat rate. In order from hot to cold their boundary temperatures are 45, 25, 10, −5 and −15°C. Rank their conductivities from lowest to highest, allowing ties. Use these explicit values if the source diagram’s left label is clipped.

Answer: I < II = III < IV.

  1. Read layer temperature drops: 20, 15, 15 and 10 K. Why: The same steady heat rate crosses equal areas and thicknesses.

  2. From heat rate = kAΔT/L, conductivity is inversely proportional to each drop. Why: The largest drop has the lowest conductivity, and equal drops give the II–III tie.

Heat Transfer · question 4

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Question: A builder needs an area-normalized insulation resistance of 35 m²·K/W and wants the layer as thin as possible. Which listed material requires the smallest thickness? Use the conductivities in Table 7.3.1.

Answer: White pine.

  1. Rearrange the required resistance to L = Rk. Why: The same target R is required for every candidate.

  2. Choose the smallest listed conductivity, white pine. Why: Lower k reaches that resistance with a thinner layer.

Heat Transfer · question 5

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Question: A cube of side 3.0 m is initially at 22°C in a large uniform 100°C enclosure. All six faces are exposed, with emissivity 0.65 and view factor 1. Ignoring conduction and convection, find its initial net radiative power into the cube. Use σ = 5.670374419 × 10⁻⁸ W/(m²·K⁴).

Answer: +23.5 kW into the cube initially.

  1. Find total exposed area A = 6(3²) = 54 m² and temperatures 373.15 K and 295.15 K. Why: Radiation uses absolute temperature and every exposed face.

  2. Use P_net,in = εσA(T_enclosure⁴ − T_cube⁴) ≈ (0.65)(5.670374419×10⁻⁸)(54)(373.15⁴ − 295.15⁴) ≈ 23.5 kW. Why: The warmer enclosure supplies net energy inward; outward emission alone would have the wrong net sign.

Latent Heat · question 1

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Question: At atmospheric pressure, 1.0 g of liquid water at 40°C receives 2.50 kJ. Use c_water = 4.1813 J/(g·K) and L_v = 2257 J/g. Assume no heat loss and retain all material. What is the final temperature?

Answer: 100.0°C.

  1. Warm the water to boiling: Q₁ = (1)(4.1813)(100 − 40) = 250.878 J. Why: Sensible heating happens before vaporization.

  2. The remaining 2249.122 J is slightly less than 2257 J needed to vaporize all 1 g. Why: Liquid and vapor coexist at 100°C while the final portion changes phase; no energy is left to superheat all the vapor.

Latent Heat · question 2

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Question: How much energy is released when 10 kg of water initially at 0°C completely freezes to ice at 0°C? Use L_f = 334 kJ/kg.

Answer: 3.34 MJ released.

  1. The temperature is unchanged through freezing. Why: A phase change uses latent heat instead of mcΔT.

  2. Find released heat magnitude Q = mL_f = (10)(334) = 3340 kJ = 3.34 MJ. Why: Heat entering the water is negative; the question asks the released magnitude.

Latent Heat · question 3

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Question: Choose the graph that correctly represents the expect relationship between temperature and heat added, if heat is added to a solid that eventually becomes a gas.

Answer: Option 4, with two phase-change plateaus.

  1. As heat is added within one phase, temperature rises. Why: The slope depends on that phase’s heat capacity.

  2. At melting and boiling, draw flat temperature intervals while heat continues to enter. Why: Latent heat changes phase without raising temperature at the stipulated constant pressure.

Latent Heat · question 4

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Question: At atmospheric pressure, will all of a 5 kg ice block initially at −10°C melt if 10 MJ is added? Use c_ice = 2.11 kJ/(kg·K) and L_f = 334 kJ/kg; ignore heat loss.

Answer: Yes.

  1. Warm the ice to 0°C: Q₁ = (5)(2.11)(10) = 105.5 kJ. Why: The ice must reach its melting point before melting completely.

  2. Melt it with Q₂ = (5)(334) = 1670 kJ; total 1775.5 kJ = 1.7755 MJ < 10 MJ. Why: The available energy exceeds both the warming and fusion requirements.

Latent Heat · question 5

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Question: Given a liquid with c = 2.05 J / g ∘ C , its gas has c = 1.5 J / g ∘ C , and the boiling point is 125 ∘ C , what is the heat of vaporization if it takes 15 MJ of energy to heat 5 kg of the liquid from 80 ∘ C to 150 ∘ C ?

Answer: About 2870 J/g.

  1. Convert mass to 5000 g and separate the heating stages. Why: The material warms as liquid, vaporizes, then warms as gas.

  2. Subtract sensible heat (5000)(2.05)(45) + (5000)(1.5)(25) = 648750 J from 15,000,000 J. Why: Only the remainder supplies latent heat.

  3. Divide by 5000 g to get L_v = 2870.25 J/g. Why: Using the full 15 MJ as latent heat would ignore both temperature changes.

Temperature · question 1

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Question: What is the temperature increase of 10 kg of water after adding 10 kJ of heat? c = 4.1813 J / g ∘ C

Answer: 0.239°C increase.

  1. Convert 10 kg to 10,000 g and 10 kJ to 10,000 J. Why: The given heat capacity uses grams.

  2. Find ΔT = Q/(mc) = 10000/(10000·4.1813) ≈ 0.2392 K. Why: Temperature differences have the same numerical value in kelvins and Celsius degrees.

Temperature · question 2

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Question: At constant pressure, 10 g of air with c_p = 1.012 J/(g·K) receives 200 J and ends at 20°C. What was its initial temperature in kelvins?

Answer: 273.4 K.

  1. Find ΔT = 200/(10·1.012) ≈ 19.763 K. Why: Constant-pressure heating uses the specified c_p.

  2. Subtract from T_f = 293.15 K to get T_i = 273.387 K. Why: Absolute temperature conversion adds 273.15 to Celsius, rather than changing ΔT.

Temperature · question 3

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Question: At constant pressure, how much heat is needed to warm 50 kg of air from 25°C to 80°C? Use c_p = 1.012 J/(g·K) and neglect losses.

Answer: 2783 kJ.

  1. Convert 50 kg to 50,000 g and find ΔT = 80 − 25 = 55 K. Why: The heat capacity is per gram per kelvin.

  2. Compute Q = (50000)(1.012)(55) = 2,783,000 J = 2783 kJ. Why: The given model neglects heat loss.

Temperature · question 4

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Question: What is the mass of a piece of lead if 250 J of energy raises its temperature 5 ∘ C ? c = 0.129 J / g ∘ C

Answer: 387.6 g.

  1. Use m = Q/(cΔT). Why: The heat supplied raises the lead’s temperature without a phase change.

  2. Calculate 250/(0.129·5) = 387.597 g. Why: The heat-capacity units determine that the answer is in grams.

Temperature · question 5

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Question: A brake pad applies a resisting force of 10 N over 15 m. All 150 J of dissipated mechanical energy enters the 50 g pad, initially at 15°C, with c = 2.01 J/(g·K). What is its final temperature?

Answer: 16.5°C.

  1. Find ΔT = 150/(50·2.01) ≈ 1.4925 K. Why: All stipulated dissipated work enters the pad as thermal energy.

  2. Add this to 15°C to get 16.4925°C. Why: A resisting force does negative mechanical work but warms the pad when that energy is dissipated there.

FFR - Thermal Expansion · question 1

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Question: A hot metal rod is placed into cool water. Indicate on the image which object in the system is gaining energy ( + Q ) and losing energy ( - Q ) .

Answer: Water gains +Q; hot rod loses −Q.

  1. Follow the initial temperature difference. Why: Heat transfers from the warmer rod to the cooler water.

  2. Use positive Q for energy entering each object. Why: The signs refer to each object’s boundary, not to a universal sign for heat.

FFR - Thermal Expansion · question 2

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Question: A cooler object (1) and warmer water (2) equilibrate in an isolated system. If Q_1 and Q_2 are signed heats entering each object, which equation expresses energy conservation?

Answer: Q₁ = −Q₂.

  1. Choose the two objects together as an isolated system. Why: No heat crosses its outer boundary.

  2. Then Q₁ + Q₂ = 0, so Q₁ = −Q₂. Why: The cooler object’s gain has the same magnitude as the warmer object’s loss.

FFR - Thermal Expansion · question 3

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Question: A silver block of 1 kg is heated to an initial temperature of 150 ∘ C . The silver block is then placed in 10 kg of cold water that has an initial temperature of 15 ∘ C . What is the difference in the final temperature that the water reaches after equilibrium is achieved? The specific heats of the silver rod and water are 0.0564 cal/g ⋅ K and 1 cal/g ⋅ K respectively.

Answer: The water warms by 0.76°C.

  1. Set (1000)(0.0564)(150 − T_f) = (10000)(1)(T_f − 15). Why: In the isolated model, silver’s heat loss equals water’s gain in matching calorie units.

  2. Find T_f = [56.4(150) + 10000(15)]/(10056.4) ≈ 15.7571°C. Why: Both materials approach a common equilibrium temperature.

  3. Subtract 15°C to get 0.7571°C. Why: The water’s much larger heat capacity keeps the rise small.

FFR - Thermal Expansion · question 4

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Question: A concrete rod has initial length 20 cm at 20°C. If its length formally increases by 3 cm, what temperature does the constant-α linear formula predict using α = 12 × 10⁻⁶ K⁻¹? Also assess whether that large extrapolation is physically valid for concrete.

Answer: 12,520°C formally; the extrapolation is not physically valid.

  1. Use ΔL = αL₀ΔT to get ΔT = 3/(12×10⁻⁶·20) = 12,500 K. Why: The centimetre ratio cancels without conversion.

  2. Add 20°C to obtain 12,520°C, then reject an intact concrete rod at that temperature. Why: Constant small-expansion coefficients cannot be extended through enormous heating and material changes.

FFR - Thermal Expansion · question 5

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Question: A diamond sphere has radius 10 cm at 25°C. If its radius formally becomes 16 cm, what temperature in Fahrenheit does the constant-α linear formula predict for α = 1.2 × 10⁻⁶ K⁻¹? Assess the validity of extrapolating the intact-solid model through this 60% expansion.

Answer: 900,077°F formally; the intact-solid model is invalid.

  1. Use radial expansion Δr/r₀ = αΔT, giving ΔT = (16 − 10)/(10·1.2×10⁻⁶) = 500,000 K. Why: The formal linear model scales each length by the same fractional change.

  2. Add 25°C and convert: T_F = 1.8(500025) + 32 = 900077°F. Why: A 60% expansion is outside the small-change intact-solid model; this is a warning about extrapolation, not a feasible diamond temperature.

Model references

These original worked steps use the stated question models. Physics references: Temperature and heat, Phase changes. AP topic labels follow the supplied-edition College Board CED, Effective Fall 2024, ©2026, pinned in the course audit.

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