Heat Transfer and Thermal Equilibrium热传递与热平衡

Big Idea · Energy (ENE)AP Science Practice 6 · ArgumentationAP link · Unit 6~40 min

Learning goals

  • Use particle collisions to explain why heat flows from a hotter object to a colder one until thermal equilibrium is reached (equal temperatures and equal average kinetic energies)
  • Use conservation of energy to show that, in an insulated system, the heat released by one part equals the heat absorbed by the other, and calculate the final temperature
  • Compare the temperature changes of two objects using their heat capacities m·c

Before you start

  • Energy, heat, and temperature (2.10.1)
  • Specific heat and q = mcΔT (2.10.3)
  • Kinetic molecular theory (U3.5)

Heat transfer at the particle level

Temperature reflects the average kinetic energy of particles (U3.5). When a hot object touches a cold one, particles at the boundary keep colliding, and on average the particles with more kinetic energy pass some of it to those with less. Energy transferred because of a temperature difference is heat (q)热.

So heat always flows from the hotter object to the colder one: the hotter object’s particles lose average kinetic energy and its temperature falls, while the colder object’s particles gain average kinetic energy and its temperature rises. This continues until the temperatures are equal, at thermal equilibrium热平衡.

At thermal equilibrium:

  • the particles of both substances have the same average kinetic energy, whether or not the particles have the same mass;
  • the particles are still moving and colliding; there is just no longer any net energy transfer between the two;
  • the particles of each substance still have a range of speeds; only the average is the same.

Conservation of energy: heat released = heat absorbed

In an insulated container no heat escapes, so:

qreleased=−qabsorbedm1c1(Tf−T1)=−m2c2(Tf−T2)q_{\text{released}} = -q_{\text{absorbed}} \qquad m_1c_1(T_{\text{f}} - T_1) = -m_2c_2(T_{\text{f}} - T_2)

Solving for the final temperature:

Tf=m1c1T1+m2c2T2m1c1+m2c2T_{\text{f}} = \frac{m_1c_1T_1 + m_2c_2T_2}{m_1c_1 + m_2c_2}

Whose temperature changes more?

Both exchange the same amount of heat, so the size of each temperature change is inversely proportional to its heat capacity热容, m·c: the one with the smaller m·c changes temperature more. When equal masses of copper and water touch, copper’s specific heat (0.385 J/(g·°C)) is only about a tenth of water’s (4.18 J/(g·°C)), so copper’s temperature changes far more, and the final temperature ends up close to the water’s starting temperature.

Choose two substances, masses, and starting temperatures to see the final temperature; then see how the energy distribution changes when two gas samples mix:

Try it: when two things touch, what is the final temperature?
Object 1
Object 2

Example 1: A metal block in water

A 50.0 g copper block at 95.0 °C is placed in 100.0 g of water at 22.0 °C in an insulated cup. Find the final temperature, and say which temperature changes more. Specific heats: copper 0.385 J/(g·°C), water 4.18 J/(g·°C) (OpenStax Table 5.1).

Step-by-step solution

  1. Conservation of energy. In an insulated cup, the heat lost by the copper equals the heat gained by the water, so their q values add up to 0 (q = mcΔT):

    qCu+qwater=0q_\text{Cu} + q_\text{water} = 0

  2. Heat capacities mc first. In J/°C:

    (mc)Cu=50.0×0.385=19.25(mc)water=100.0×4.18=418\begin{aligned} (mc)_\text{Cu} &= 50.0 \times 0.385 = 19.25 \\ (mc)_\text{water} &= 100.0 \times 4.18 = 418 \end{aligned}

  3. Substitute and solve for the final temperature T.

    19.25(T−95.0)+418(T−22.0)=0\begin{aligned} &19.25(T - 95.0) \\ &+ 418(T - 22.0) = 0 \end{aligned} T=19.25×95.0+418×22.019.25+418=25.2 ∘C\begin{aligned} T &= \frac{19.25 \times 95.0 + 418 \times 22.0}{19.25 + 418} \\ &= \boxed{25.2\ ^\circ\text{C}} \end{aligned}

  4. Compare the temperature changes.

    Cu:95.0−25.2=69.8water:25.2−22.0=3.2\begin{aligned} \text{Cu} &: 95.0 - 25.2 = 69.8 \\ \text{water} &: 25.2 - 22.0 = 3.2 \end{aligned}

    The copper’s temperature changes far more.

  5. Check and explain. They exchange the same heat (in J):

    19.25×69.8=1.34×103418×3.2=1.34×103\begin{aligned} 19.25 \times 69.8 &= 1.34 \times 10^3 \\ 418 \times 3.2 &= 1.34 \times 10^3 \end{aligned}

    Same heat, but copper’s heat capacity is only about 1/22 of the water’s, so its temperature changes much more.

Example 2: Mixing two samples of water

60.0 g of water at 50.0 °C is mixed with 40.0 g of water at 15.0 °C in an insulated cup. Find the final temperature.

Step-by-step solution

  1. Conservation of energy. Heat lost by the hot water = heat gained by the cold water:

    60.0c(T−50.0)+40.0c(T−15.0)=0\begin{aligned} &60.0c(T - 50.0) \\ &+ 40.0c(T - 15.0) = 0 \end{aligned}

  2. Cancel the specific heat. It’s the same substance, so c is the same and cancels:

    60.0(T−50.0)+40.0(T−15.0)=0\begin{aligned} &60.0(T - 50.0) \\ &+ 40.0(T - 15.0) = 0 \end{aligned}

  3. Solve for T.

    T=60.0×50.0+40.0×15.0100.0=36.0 ∘C\begin{aligned} T &= \frac{60.0 \times 50.0 + 40.0 \times 15.0}{100.0} \\ &= \boxed{36.0\ ^\circ\text{C}} \end{aligned}

    This is just the mass-weighted average temperature.

  4. Is it reasonable? 36.0 °C lies between 15.0 °C and 50.0 °C, closer to the temperature of the larger, hot sample.

  5. Check the heat. In J:

    lost=60.0×4.18×14.0=3.51×103gained=40.0×4.18×21.0=3.51×103\begin{aligned} \text{lost} &= 60.0 \times 4.18 \times 14.0 \\ &= 3.51 \times 10^3 \\ \text{gained} &= 40.0 \times 4.18 \times 21.0 \\ &= 3.51 \times 10^3 \end{aligned}

    They’re equal.

Practice: two objects reach thermal equilibrium in an insulated container. Whose temperature changes more?
  1. A: 10 g of copper; B: 10 g of water
  2. A: 100 g of aluminum; B: 100 g of gold
  3. A: 10 g of water; B: 20 g of water
  4. A: 50 g of water; B: 50 g of ethanol
  5. A: 20 g of water; B: 20 g of water
  6. A: 200 g of iron (m·c = 89.8 J/°C); B: 20 g of water (m·c = 83.6 J/°C)
  7. A: 100 g of copper; B: 10 g of copper
  8. A: 50 g of iron; B: 50 g of aluminum

Common mistakes

Mistake 1: “Whichever changes temperature more exchanges more heat.” Both exchange the same amount of heat. A larger temperature change only means a smaller heat capacity m·c.

Mistake 2: “At thermal equilibrium, particles stop moving (or stop colliding).” Particles keep moving and colliding. Thermal equilibrium only means there is no net heat transfer between the two.

Mistake 3: “At thermal equilibrium, heavier atoms have more average kinetic energy.” At the same temperature the average kinetic energy is the same, whatever the particles’ masses. Heavier particles just have a lower average speed.

Quick check

A hot iron block is placed in cold water in an insulated container until thermal equilibrium is reached. Which statement is correct?

80.0 g of water at 30.0 °C is mixed with 20.0 g of water at 70.0 °C (insulated). What is the final temperature?

A 200 g iron pan and 200 g of water each absorb 5.0 kJ of heat from a stove (no phase change). Which warms more?

Equal amounts of argon at 300 K and at 500 K are mixed in an insulated container. At thermal equilibrium:

Heat flows from hot water to cold water. Which is the best particle-level explanation?

A 100 g copper block and a 10 g aluminum block, both at 50 °C, are pressed together (insulated). What happens?

AP-style free response

AP-style free response

To identify a metal, a student heats a 75.0 g block in boiling water to 100.0 °C, then quickly places it in 150.0 g of water at 21.0 °C in an insulated cup. At thermal equilibrium the temperature is 25.0 °C.

(a) Which way does heat flow? Explain why at the particle level.

(b) Calculate the heat absorbed by the water (c(water) = 4.18 J/(g·°C)).

(c) Calculate the specific heat of the metal.

(d) Using a table of specific heats (aluminum 0.897, iron 0.449, copper 0.385, gold 0.129 J/(g·°C)), which metal is it most likely to be?

(e) Moving the block from the boiling water to the cup took a few seconds, so it was actually only about 97 °C when it entered the water, but the student still used 100.0 °C. Would the calculated specific heat be too large or too small? Explain.

Model answer and scoring

(a) From the metal to the water. The metal’s atoms have more average kinetic energy, and in collisions they pass energy to the water molecules until the temperatures are equal.

(b) q(water) = 150.0 g × 4.18 J/(g·°C) × (25.0 − 21.0) °C = 2.51 × 10³ J.

(c) The metal releases 2.51 × 10³ J: c = 2508 J ÷ (75.0 g × 75.0 °C) = 0.446 J/(g·°C).

(d) Iron (0.449 J/(g·°C)).

(e) Too small. The heat absorbed by the water was measured correctly, but the metal’s actual temperature drop (about 97 − 25.0 °C) was smaller than the 75.0 °C the student used; dividing the same heat by too large a ΔT gives too small a specific heat.

Scoring: (a) direction and energy transfer by collisions; (b) q(water); (c) uses q released = q absorbed to find c; (d) the identification; (e) too small, with a reason.

Summary

  • Heat passes from a hotter object to a colder one through particle collisions until the temperatures are equal (thermal equilibrium); at equilibrium the average kinetic energies are equal, and particles keep moving and colliding.
  • In an insulated system, the heat released by one part equals the heat absorbed by the other: m₁c₁(Tf − T₁) = −m₂c₂(Tf − T₂).
  • When the same heat is exchanged, the object with the smaller heat capacity m·c changes temperature more.
  • The final temperature is the average of the starting temperatures weighted by m·c, so it lies between them.

AP tip: To compare temperature changes, compare m·c, not just specific heat or just mass. To explain thermal equilibrium, say “equal average kinetic energy and no net energy transfer,” not that particles stop moving.

Practice this chapter: Unit 6 · Thermochemistry · 12 questionsPractice this topic: U6.3 · Heat Transfer and Thermal Equilibrium · 6 questions

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