How much heat is released when 105 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) 273 kJ C) 319 kJ D) 347 kJ

Question

Quizplus · Accepted Answer

First, we need to determine how much energy is released when the steam is cooled to 100°C water:

q1 = m * Cp * ΔT 
q1 = (105 g / 18.015 g/mol) * (75.4 J/mol°C) * (100°C - 100°C) 
q1 = 0 J

No energy is released when steam condenses to 100°C water since there is no temperature change.

Next, we need to determine how much energy is released when the water is cooled to 0°C:

q2 = m * ΔHfus 
q2 = (105 g / 18.015 g/mol) * (6.01 kJ/mol) 
q2 = 35.1 kJ

Now, we need to determine how much energy is released when the water at 0°C is cooled to -15°C:

q3 = m * Cp * ΔT 
q3 = (105 g / 18.015 g/mol) * (36.4 J/mol°C) * (-15°C - 0°C) 
q3 = -22.5 kJ

Since energy is released, the value is negative.

Finally, we need to determine how much energy is released when the ice is formed from the water:

q4 = m * ΔHfus 
q4 = (105 g / 18.015 g/mol) * (6.01 kJ/mol) 
q4 = 35.1 kJ

The total energy released is the sum of q1, q2, q3, and q4:

ΔH = q1 + q2 + q3 + q4 
ΔH = 0 J + 35.1 kJ - 22.5 kJ + 35.1 kJ 
ΔH = 47.7 kJ

However, we need to account for the fact that this energy is released per mole of water, not per gram of water. To convert to the correct units:

ΔH = (47.7 kJ / mol) * (18.015 g/mol) 
ΔH = 858.3 J/g

Therefore, the correct answer is C) 319 kJ.

How Much Heat Is Released When 105 G of Steam