U6.4 · Heat Capacity and CalorimetryPractice · 热容与量热
Lessons covered:U6.4 Heat Capacity and Calorimetry
- 1Multiple choice
Two pots are made of the same stainless steel: a large one of 2.0 kg and a small one of 0.50 kg. Which statement is correct?
ExplanationSpecific heat c is a property of the material (per gram), so it is the same for both. Heat capacity C = m·c belongs to the whole object, so 4 times the mass gives 4 times the heat capacity. For the same heat, ΔT = q ÷ C, so the large pot warms only 1/4 as much.
- 2Numeric
In a coffee-cup calorimeter, 2.12 g of (0.0500 mol) dissolves in 100.0 g of water, and the temperature rises from 21.5 °C to 25.8 °C. Take the solution’s specific heat as 4.18 J/(g·°C) and ignore heat absorbed by the calorimeter. What is the enthalpy of solution of LiCl, in kJ/mol?
kJ/molExplanationMass of solution = 2.12 + 100.0 = 102.1 g; qsoln = 102.1 × 4.18 × 4.3 = 1836 J. The solution warmed, so dissolving is exothermic: ΔH = −1.836 kJ ÷ 0.0500 mol = −36.7 kJ/mol.
- 3Multiple choice
In measuring the enthalpy of solution of LiCl, the LiCl used had been sitting out and had absorbed some water from the air. The student still treats the whole weighed mass as LiCl when calculating moles. How does this affect the magnitude of the calculated enthalpy of solution?
ExplanationPart of the weighed mass is water, so less LiCl actually dissolves and the heat q comes out smaller, while the n used in the calculation still assumes it is all LiCl and is too large. |ΔH| = q ÷ n has a smaller numerator and a larger denominator, so the result is too small.
- 4Numeric
Aluminum has a specific heat of 0.897 J/(g·°C) and a molar mass of 26.98 g/mol. What is its molar heat capacity?
J/(mol·°C)ExplanationMolar heat capacity is the heat needed to raise 1 mol by 1 °C: 0.897 J/(g·°C) × 26.98 g/mol = 24.2 J/(mol·°C). The grams cancel.
- 5Numeric
A reaction is run in a calorimeter holding 50.0 g of solution with a specific heat of 4.18 J/(g·°C). The calorimeter itself has a heat capacity of 12.0 J/°C. The temperature rises by 3.20 °C. How much heat did the reaction release?
JExplanationBoth the solution and the calorimeter warm up, so both absorbed the reaction’s heat: q = (m·c + Ccal) × ΔT = (50.0 × 4.18 + 12.0) × 3.20 = 221.0 × 3.20 = 707 J. Counting only the solution gives a too-small 669 J.
- 6Multiple choice
In a coffee-cup calorimeter, 5.00 g of is added to 100.0 g of water at 1 min, and the temperature is recorded. The calorimeter slowly loses heat, so the later cooling line (dashed) is extrapolated back to 1 min. Which temperature change ΔT best represents the heat released by dissolving?
ExplanationWith no heat loss, the temperature would jump to its maximum the moment the salt dissolves; in reality heat leaks away while it rises, so the highest reading (about 28.9 °C) is too low. Extrapolating the cooling line back to the moment of mixing (1 min) gives about 29.3 °C, so ΔT = 29.3 − 21.0 ≈ 8.3 °C. Check: 0.0451 mol × 81.3 kJ/mol ÷ (105.0 g × 4.18) ≈ 8.3 °C. 29.3 °C is a temperature, not a change.
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