Question

One kg of a steam with quality of 20 percent is heated at a constant pressure of 200kPa until the temperature reaches 400°C. Calculate the work done by the system.

Answer 1

$$W=mP(v_2-v_1)$$

From steam tables:

$$v_1=0.8857\frac{m^3}{kg}$$

$$v_2=1.5493\frac{m^3}{kg}$$

$$W=(1)(200000)(1.5493-0.8857)=132700\ J=132.7\ kJ$$