Question

(a). Suppose 384 g of steam (i.e., water vapor) originally at 100°C is quickly cooled to produce liquid water at 31°C. How much heat must be removed from the steam to accomplish this? Some of the following data may be helpful. Cp(H2O(g)) = 36.4 J mol–1 K–1 Tboil(H2O) = 100°C Cp(H2O(!)) = 75.3 J mol–1 K–1 ∆H°vaporization(H2O) = 40.7 kJ mol–1 (8 pts) (b). Use information given above to estimate the entropy change for the vaporization of water (i.e., ∆S° for the process H2O(!) ® H2O(g))

Answer 1

Answer:

a) -978 kJ

b) -2.32 kJ/K

Explanation:

The molar mass of water is 18.02 g/mol. The moles represented by 384 g are:

384 g × (1 mol/ 18.02 g) = 21.3 mol

a) The total heat lost by the steam is the sum of the heats lost in 2 stages.

1) Vapor to liquid water at 100 °C (condensation)

Q₁ = -ΔH°vap × n = - (40.7 kJ/mol) × 21.3 mol = -867 kJ

2) Liquid water: from 100°C (373 K) to 31 °C (304 K)

Q₂ = c(l) × n × ΔT = (75.3 J/mol.K) × 21.3 mol × (304 K - 373 K) = -1.11 × 10⁵ J = -111 kJ

Total heat = Q₁ + Q₂ = -867 kJ + (-111 kJ) = -978 kJ

b) The entropy change for the vaporization can be calculated using the following expression.

ΔS = Q₁ / Tboil = -867 kJ / 373 K = -2.32 kJ/K