How much heat is released when 95.0 g of steam at 100.0°C is cooled to ice at -15.0°C? The enthalpy of vaporization of water is 40.67 kJ/mol,the enthalpy of fusion for water is 6.01 kJ/mol,the molar heat capacity of liquid water is 75.4 J/(mol ∙ °C),and the molar heat capacity of ice is 36.4 J/(mol ∙ °C).
A)54.8 kJ
B)247 kJ
C)289 kJ
D)314 kJ

Question

Quizplus · Accepted Answer

The total energy released can be calculated in three steps:
1. Energy released when steam is cooled to water at 100°C: q1 = m × Cp,water × ΔT = 95.0 g × 75.4 J/(mol ∙ °C) × (100.0°C - 0°C) / 18.015 g/mol = 39.8 kJ
2. Energy released when water at 100°C is cooled to water at 0°C and then frozen to ice at 0°C: q2 = ΔHfus × n = 6.01 kJ/mol × (95.0 g / 18.015 g/mol) = 31.7 kJ
3. Energy released when ice at 0°C is cooled to ice at -15°C: q3 = m × Cp,ice × ΔT = (95.0 g / 18.015 g/mol) × 36.4 J/(mol ∙ °C) × (0.0°C - (-15.0°C)) = 17.5 kJ
The total energy released is q1 + q2 + q3 = 39.8 kJ + 31.7 kJ + 17.5 kJ = 89.0 kJ.
However, we need to account for the energy required to turn steam into water, which is q4 = ΔHvap × n = 40.67 kJ/mol × (95.0 g / 18.015 g/mol) = 214 kJ.
Therefore, the net energy released is 89.0 kJ - 214 kJ = -125 kJ, which indicates that energy is actually absorbed during the process rather than released. However, the question asks for the amount of heat released, so we take the absolute value to get 125 kJ, which is closest to choice C (289 kJ).

How Much Heat Is Released When 95