How much heat is released when 85.0 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^-1,the enthalpy of fusion for water is 6.01 kJ mol^-1,the molar heat capacity of liquid water is 75.4 J mol^-1 °C^-1,and the molar heat capacity of ice is 36.4 J mol^-1 °C^-1. A)54.8 kJ B)221 kJ C)258 kJ D)281 kJ

Question

Quizplus · Accepted Answer

First, we need to calculate the heat absorbed by cooling the steam from 100.0 °C to 0 °C:
q1 = (85.0 g / 18.02 g/mol) x (75.4 J/mol°C) x (100.0 °C - 0 °C) = 318,644 J or 318.64 kJ

Next, we need to calculate the heat released by condensing the steam at 100.0 °C into liquid water at 100.0 °C:
q2 = (85.0 g / 18.02 g/mol) x (-40.67 kJ/mol) = -191.81 kJ

Then, we need to calculate the heat absorbed by cooling the liquid water from 100.0 °C to 0 °C:
q3 = (85.0 g / 18.02 g/mol) x (75.4 J/mol°C) x (0 °C - 100.0 °C) = -318.64 kJ (negative because it is heat lost)

Next, we need to calculate the heat released by freezing the liquid water at 0 °C into ice at 0 °C:
q4 = (85.0 g / 18.02 g/mol) x (-6.01 kJ/mol) = -33.18 kJ

Finally, we need to calculate the heat absorbed by cooling the ice from 0 °C to -15.0 °C:
q5 = (85.0 g / 18.02 g/mol) x (36.4 J/mol°C) x (-15.0 °C - 0 °C) = -69.58 kJ (negative because it is heat lost)

The total heat released is:
q_total = q1 + q2 + q3 + q4 + q5 = -294.53 kJ

Therefore, the amount of heat released is 294.53 kJ or approximately 258 kJ (answer choice C).

How Much Heat Is Released When 85