Question

1. 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. a) 10.2 kj b) 3.2 kj c) 15.6 kj d) 5.4 kj e) 1.6 kj

Explanation:

Step1: Cool vapor to 100°C

Heat lost: \( q_1 = m \cdot c_{\text{vapor}} \cdot \Delta T \)
\( m = 5.00 \, \text{g}, c_{\text{vapor}} = 1.84 \, \text{J/g·K}, \Delta T = 155 - 100 = 55 \, \text{K} \)
\( q_1 = 5.00 \times 1.84 \times 55 = 506 \, \text{J} \)

Step2: Condense vapor to liquid

Heat lost: \( q_2 = m \cdot \Delta H_{\text{vaporization}} \)
\( \Delta H_{\text{vaporization}} = 2260 \, \text{J/g} \)
\( q_2 = 5.00 \times 2260 = 11300 \, \text{J} \)

Step3: Cool liquid to 0°C

Heat lost: \( q_3 = m \cdot c_{\text{liquid}} \cdot \Delta T \)
\( c_{\text{liquid}} = 4.18 \, \text{J/g·K}, \Delta T = 100 - 0 = 100 \, \text{K} \)
\( q_3 = 5.00 \times 4.18 \times 100 = 2090 \, \text{J} \)

Step4: Freeze liquid to ice

Heat lost: \( q_4 = m \cdot \Delta H_{\text{fusion}} \)
\( \Delta H_{\text{fusion}} = 336 \, \text{J/g} \)
\( q_4 = 5.00 \times 336 = 1680 \, \text{J} \)

Step5: Cool ice to -55°C

Heat lost: \( q_5 = m \cdot c_{\text{ice}} \cdot \Delta T \)
\( c_{\text{ice}} = 2.09 \, \text{J/g·K}, \Delta T = 0 - (-55) = 55 \, \text{K} \)
\( q_5 = 5.00 \times 2.09 \times 55 = 574.75 \, \text{J} \)

Step6: Total heat lost

Total \( q = q_1 + q_2 + q_3 + q_4 + q_5 \)
\( q = 506 + 11300 + 2090 + 1680 + 574.75 = 15650.75 \, \text{J} = 15.65 \, \text{kJ} \approx 15.6 \, \text{kJ} \)

Answer:

C) 15.6 kJ