Answer
a) $T_3=\frac{1}{(\dot{m}_1+\dot{m}_2)}(\frac{\dot{Q}}{c_p}+\dot{m}_1T_1+\dot{m}_2T_2)$
b) $\dot{V}_3=\frac{\dot{R.Q}}{P.c_p}+\dot{V}_1+\dot{V}_2$
c) $\dot{V}_3=\dot{V}_1+\dot{V}_2$
Work Step by Step
a) Taking $T_{ref}=0K \rightarrow h=c_p(T-T_{ref})=c_pT$
The energy balance is:
$\dot{Q}+\dot{m}_1h_1+\dot{m}_2h_2=(\dot{m}_1+\dot{m}_2)h_3$
$\dot{Q}+\dot{m}_1c_pT_1+\dot{m}_2c_pT_2=(\dot{m}_1+\dot{m}_2)c_pT_3$
$T_3=\frac{1}{(\dot{m}_1+\dot{m}_2)}(\frac{\dot{Q}}{c_p}+\dot{m}_1T_1+\dot{m}_2T_2)$
b) $\dot{V}_3=\frac{\dot{m}_3RT_3}{P_3}$
$\dot{V}_3=\frac{\dot{m}_3R}{P_3}\times\frac{1}{\dot{m}_3}(\frac{\dot{Q}}{c_p}+\dot{m}_1T_1+\dot{m}_2T_2)$
Since $P_1=P_2=P_3$
$\dot{V}_3=\frac{\dot{R.Q}}{P.c_p}+\frac{\dot{m}_1RT_1}{P_1}+\frac{\dot{m}_21RT_2}{P_2}$
$\dot{V}_3=\frac{\dot{R.Q}}{P.c_p}+\dot{V}_1+\dot{V}_2$
c) If $\dot{Q}=0$:
$\dot{V}_3=\dot{V}_1+\dot{V}_2$
