Explanation:
[tex]\begin{array}{ll}\1 & \2 \\P_{1}=0-\_{\text {mpa }} \text { Steam } & P_{2}=0.06 mpq \\x=0.9 & x=0 \text { (satunated liquid) }\end{array}[/tex]
a) Term of entering steam will be Tsaturated of corresponding pressure
[tex]so, At P =0.06 mpa , T_{s a t}=85.92^{\circ} C[/tex]
b)
[tex]\begin{aligned}Q &=m c_{p}\left(T_{0}-T_{i}\right) \\&=\frac{100}{60} \times 1.005(60-30)=50.25 W\end{aligned}[/tex]