Answer
$T_2=46.0°C$
Work Step by Step
The flowrate is given by:
$\dot{m}=\frac{P_1\dot{V}_1}{RT_1}$
Given $P_1=101.325\ kPa,\ \dot{V}_1=0.0008\ m³/s,\ R=0.287\ kJ/kg.K,\ T_1=25°C$:
$\dot{m}=0.0009477\ kg/s$
From the energy balance:
$\dot{Q}+\dot{m}h_1=\dot{m}h_2$
$\dot{Q}=\dot{m}c_p(T_2-T_1)$
and since $\dot{Q}=15W,\ c_p=1.005\ kJ/kg.C$:
$T_2=46.0°C$
