Answer
$\dot{Q}_s=63\ W$
Work Step by Step
The flowrate is given by:
$\dot{m}=\frac{P_1\dot{V}_1}{RT_1}$
Given $P_1=101.325\ kPa,\ \dot{V}_1=0.6\ m³/min,\ R=0.287\ kJ/kg.K,\ T_1=30°C$:
$\dot{m}=0.70\ kg/min$
From the energy balance:
$\dot{Q}_a+\dot{m}h_1=\dot{m}h_2$
$\dot{Q}_a=\dot{m}c_p(T_2-T_1)$
and since $T_2=40°C,\ c_p=1.005\ kJ/kg.C$:
$\dot{Q}_a=0.117\ kW=117\ W$
Therefore: $\dot{Q}=\dot{Q}_a+\dot{Q}_s,\ \dot{Q}=180W$:
$\dot{Q}_s=63\ W$
