Question

Exactly 1.73 kg of water vapor is contained in a piston–cylinder assembly at a pressure of
1.00 MPa and temperature 600. °C. The vapor is isothermally compressed to 80.0 MPa.
Determine the sum of the work and heat energy transports in this process.

Answer 1

Question #79172

Exactly $1.73\,\mathrm{kg}$ of water vapor is contained in a piston-cylinder assembly at a pressure of $1.00\,\mathrm{MPa}$ and temperature $600^{\circ}\mathrm{C}$. The vapor is isothermally compressed to $80.0\,\mathrm{MPa}$.

Determine the sum of the work and heat energy transports in this process.

Answer:

The first law of thermodynamics is given by:

$$\Delta U = Q - W,$$

where $\Delta U$ is the change in the internal energy during a process,

$Q$ and $W$ are the heat and work transport, respectively.

Since there is no change in the internal energy during an isothermal process, (1) gives:

$$Q = W,$$

The heat transfer during an isothermal process is given by:

$$Q = mT(s_2 - s_1),$$

where $m = 1.73\,\mathrm{kg}$ – the mass of the steam,

$T = 600^{\circ}\mathrm{C} + 273 = 873\,\mathrm{K}$ – the absolute temperature of the steam,

$s_1$ and $s_2$ – the specific entropy of the steam at the beginning and the end of the process.

From the steam table we can find the specific entropy of the steam (see https://www.nist.gov/sites/default/files/documents/srd/NISTIR5078-Tab3.pdf):

$$s_1 = 8.0310\,\mathrm{kJ/kg\cdot K} \text{ at } p_1 = 1.00\,\mathrm{MPa} \text{ and } t_1 = 600^{\circ}\mathrm{C},$$ $$s_2 = 5.3674\,\mathrm{kJ/kg\cdot K} \text{ at } p_2 = 80.0\,\mathrm{MPa} \text{ and } t_2 = t_1 = 600^{\circ}\mathrm{C}.$$

Substitute into (3):

$$Q = 1.73 \cdot 873 \cdot (5.3674 - 5.3674) = -4023\,\mathrm{kJ}.$$

Taking into account (2), we can find the sum of the work and heat energy transports in this process as a double heat:

$$Q + W = 2Q = 2 \cdot (-4023) = -8046\,\mathrm{kJ}.$$