Question

how much heat would be required to transform 2 kg of saturated liquid-vapor mixture of water at 60% quality and 80 degree celcius to saturated vapour at constant pressure and then to steam at 150 degree celcius and 0.2 MPa at constant volume?

Answer 1

Gives

M=2kg

$∆m=2\times\frac{60}{100}=1.2kg$

$T_1=80°c$

$T_2=150°c$

P=0.2MPa

Q=∆ms ∆T

For steam

$s=1.996KJ/kgK$

$$∆Q=1.2\times1.996(150°-80°)=167.664KJ$$