Answer
$T_2=386.8\ K$
$m_2=9460\ kg$
Work Step by Step
Mass balance for air:
$\frac{dm_a}{dt}=\dot{m}_a$
Mass balance for water:
$\frac{dm_w}{dt}=-\dot{m}_w$
Energy balance for the system:
$\dot{m}_ah_a-\dot{m}_wh_w=\frac{d(mu)_a}{dt}+\frac{d(mu)_w}{dt}$
$\frac{dm_a}{dt}h_a+\frac{dm_w}{dt}h_w=\frac{d(mu)_a}{dt}+\frac{d(mu)_w}{dt}$
Integrating over time:
$[(mu)_2-(mu)_1]_a+[(mu)_2-(mu)_1]_w=h_a(m_2-m_1)_a+h_w(m_2-m_1)_w$
$\frac{PV_2}{RT_2}c_vT_2-\frac{PV_1}{RT_1}c_vT_1+0-m_{w,1}c_wT_w=c_pT_{in}(\frac{PV_2}{RT_2}-\frac{PV_1}{RT_1})+0-m_{w,1}c_wT_w$
$V_2-V_1=kT_{in}(\frac{V_2}{T_2}-\frac{V_1}{T_1})$
Given $k=1.4,\ T_{in}=293\ K,\ V_2=700\ m³,\ V_1=100\ m³,\ T_1=288\ K$
$T_2=386.8\ K$
$m_2=PV_2/RT_2,\quad P=1500\ kPa,\ R=0.287\ kJ/kg.K$
$m_2=9460\ kg$
