A 5.00 g sample of water vapor, initially at 155°C is cooled at atmospheric pressure, producing ice at −55°C. Calculate the amount of heat energy lost by the water sample in this process, in kJ. Use the following data: specific heat capacity of ice is 2.09 J/g·K; specific heat capacity of liquid water is 4.18 J/g·K; specific heat capacity of water vapor is 1.84 J/g·K; heat of fusion of ice is 336 J/g; heat of vaporization of water is 2260 J/g.

Question

Quizplus · Accepted Answer

The process can be divided into three steps: 
1. Cooling the water vapor from 155°C to 0°C and condensing it into liquid water. 
2. Cooling the liquid water from 0°C to -55°C and solidifying it into ice. 
3. Melting the ice at -55°C to 0°C.

Step 1: 
q = m × c × ΔT 
q = 5.00 g × 1.84 J/g·K × (0°C - 155°C) = -1550 J 
The negative sign indicates that energy is lost.

Step 2: 
q = m × c × ΔT 
q = 5.00 g × 4.18 J/g·K × (0°C - (-55°C)) = 1158.5 J 
The positive sign indicates that energy is gained.

Step 3: 
q = m × ΔH_fus 
q = 5.00 g × 336 J/g = 1680 J 
The positive sign indicates that energy is gained.

The total energy lost by the water sample is: 
q_total = q1 + q2 + q3 = -1550 J + 1158.5 J + 1680 J = 288.5 J

To convert J to kJ, divide by 1000: 
q_total = 0.2885 kJ

Therefore, the amount of heat energy lost by the water sample in this process is 15.6 kJ (rounded to one decimal place).

A 500 G Sample of Water Vapor, Initially at 155°C Is