Answer
$T_2=49.2°C$
Work Step by Step
At the initial state for air:
$V=4\times5\times6=120m³$
For an ideal gas (given $P_1=95\ kPa,\ R=0.2870 kJ/kg.K,\ T_1=288\ K$:
$PV=mRT$
Solving for the mass: $m_{air}=137.9\ kg$
The energy balance fr the system is:
$\Delta U_{air}+\Delta U_{water}=0$
Given $c_{v,water}=4.18\ kJ/kg.K,\ c_{v,air}=0.718\ kJ/kg.K,\ m_{water}=1000\ kg$,
$T_{1,air}=15°C, T_{1,water} = 50°C$
and solving for the final temperature:
$T_2=49.2°C$
