U6.3 · Heat Transfer and Thermal EquilibriumPractice · 热传递与热平衡
Lessons covered:U6.3 Heat Transfer and Thermal EquilibriumU6.4 Heat Capacity and Calorimetry
- 1Multiple choice
A hot piece of iron is dropped into cold water. At the particle level, how is heat transferred?
ExplanationTemperature reflects the average kinetic energy of the particles. The fast-vibrating iron atoms hand some kinetic energy to the slower water molecules when they collide at the surface; the water speeds up and the iron slows down until the average kinetic energies (temperatures) match, which is thermal equilibrium. At that point the temperatures are equal, not the amounts of energy.
- 2Numeric
A 25.0 g gold ring at 90.0 °C is dropped into 30.0 g of ethanol at 18.0 °C in an insulated cup. Specific heats: gold 0.129 J/(g·°C), ethanol 2.38 J/(g·°C). What is the final temperature?
°CExplanationHeat lost by the gold = heat gained by the ethanol: 3.225(90.0 − T) = 71.4(T − 18.0), so T = 21.1 °C. Gold’s m·c is only 3.225 J/°C against ethanol’s 71.4 J/°C, so the ring cools by 68.9 °C while the ethanol warms by just 3.1 °C.
- 3Multiple choice
A hot piece of metal is placed in cold water in an insulated cup. Once thermal equilibrium is reached, which quantity must be the same for the metal and the water?
ExplanationThermal equilibrium means equal temperatures, so the average kinetic energies match and there is no more net heat flow. |ΔT| depends on each object’s m·c, total thermal energy also depends on mass and substance, and specific heat is a property of each material that contact doesn’t change.
- 4Numeric
A 45.0 g copper block is heated in an oven and quickly dropped into 75.0 g of water at 21.0 °C in an insulated cup. The final temperature is 25.6 °C. Specific heats: copper 0.385 J/(g·°C), water 4.18 J/(g·°C). What was the copper’s temperature before it went into the water?
°CExplanationHeat absorbed by the water: 75.0 × 4.18 × (25.6 − 21.0) = 1442 J, which equals the heat the copper lost. The copper cooled by 1442 ÷ (45.0 × 0.385) = 83.2 °C, so it started at 25.6 + 83.2 = 108.8 °C.
- 5Multiple choice
A 100 g hot iron block is placed in 100 g of cold water in an insulated cup. At equilibrium, the iron’s temperature has changed much more than the water’s. What is the main reason?
ExplanationWith insulation, q lost = q gained, and ΔT = q ÷ (m·c). Iron’s m·c = 100 × 0.449 = 44.9 J/°C against water’s 418 J/°C, about 9 times larger, so the iron’s ΔT is about 9 times the water’s. How well iron conducts affects only how fast equilibrium comes, not the final temperature, and the masses are equal.
- 6Multiple choice
A 50.0 g aluminum block at 90.0 °C is placed in 100.0 g of water at 20.0 °C in an insulated container. The graph shows both temperatures against time. Which best explains why the water’s temperature changes much less than the aluminum’s?
ExplanationThe two curves meet at about 26.8 °C (thermal equilibrium). The aluminum falls 63.2 °C and the water rises 6.8 °C. With no loss, heat lost by Al = heat gained by water: 50.0 × 0.897 × 63.2 ≈ 100.0 × 4.18 × 6.8 ≈ 2835 J. The water’s heat capacity, 418 J/°C, is about 9 times the aluminum’s (44.9 J/°C), so the same heat changes its temperature only about 1/9 as much. Heat flows from the hot aluminum to the water.
Review: U6.3 Heat Transfer and Thermal EquilibriumU6.4 Heat Capacity and Calorimetry
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