Question 1.28: Use the data in the steam tables to answer the following: A)...

A)

\Delta \mathrm{U}=\operatorname{Mass}\left(\underline{\mathrm{U}}_{2}-\underline{\mathrm{U}}_{1}\right)

Steam at 100°C and 1 bar → 2506.2 \frac{\mathrm{kJ}}{\mathrm{kg}}

Steam at 300°C and 1 bar → 2810.6 \frac{\mathrm{kJ}}{\mathrm{kg}}

\Delta \mathrm{U}=(100 \mathrm{~kg})\left(2506.2 \frac{\mathrm{kJ}}{\mathrm{kg}}-2810.6 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)=-\mathbf{3 0, 440} \mathbf{~kJ}

B)

\Delta \mathrm{U}=\operatorname{Mass}\left(\underline{\mathrm{U}}_{2}-\underline{\mathrm{U}}_{1}\right)

Water at 300°C and 200 bar → 1307.2 \frac{\mathrm{kJ}}{\mathrm{kg}}

Water at 240°C and 200 bar → 1016.1 \frac{\mathrm{kJ}}{\mathrm{kg}}

\Delta \mathrm{U}=(100 \mathrm{~kg})\left(1307.2 \frac{\mathrm{kJ}}{\mathrm{kg}}-1016.1 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)=\bf 29110 \mathbf{~kJ}

C) The large temperature changes can be accomplished largely or entirely with heat transfers. In part A, some work would also be transferred if for example this process was occurring in a closed system; since the temperature of the gas is changing at constant pressure the volume must also be changing. The change in volume of the liquid in part B, however, is likely negligible.

D) No, because heat transfer requires a temperature driving force. The steam cools all the way to 100°C, and there is no way it can transfer heat to the liquid after it has fallen below 240°C.