Answer
$\Delta t = 30.82\ min$
Work Step by Step
Energy balance for the whole system:
$Q_{rad}-Q_{lost} = \Delta U_{air} + \Delta H_{oil}$
With:
$Q_{rad} = 2.4\ kW\times \Delta t,\ Q_{lost} = 0.35\ kW\times \Delta t$
$ \Delta U_{air} = m_{air}c_{v,air}\Delta T_{air},\ \Delta H_{oil} = m_{oil}c_{p,oil}\Delta T_{oil}$
For the oil $V=40\ L,\ \rho=950\ kg/m³,\ c_p=2.2\ kJ/kg.°C,\ \Delta T=40°C$:
$m = 38 kg$
$\Delta H_{oil} = 3344\ kJ$
For the air $R=0.2870 kJ/kg.K,\ P_1=101.325\ kPa,\ V=50\ m³$,
$T_1=10°C=283.15\ K, c_v=0.718\ kJ/kg.°C,\ \Delta T_{air}=10°C$:
$m_{air}=\frac{PV}{RT} = 62.34\ kg$
$ \Delta U_{air}=447.62\ kJ$
Therefore:
$\Delta t = \frac{3791.6\ kJ}{2.05\ kW} = 1849.56\ s\times \frac{1\ min}{60\ s}$
$\Delta t = 30.82\ min$
