Answer
a) $\dot{Q}=5.17\ kW$
b) $\Delta T=-6.9°C$
Work Step by Step
Given data $h_1=220.2\ kJ/kg,\ v_1 = 0.0253\ m3/kg,\ h_2= 398.0\ kJ/kg$
For the refrigerant ($\dot{V}_1=2.65\ L/h$):
$\dot{m}_1=\dot{V}_1/v_1$
$\dot{m}_1=0.0291\ kg/s$
$\dot{Q}+\dot{m}_1h_1=\dot{m}_1h_2$
So $\dot{Q}=5.17\ kW$
For the air $\dot{m}_3=0.75\ kg/s,\ c_p=1.005\ kJ/kg.K$:
$\dot{Q}=-\dot{m}_3c_p\Delta T$
$\Delta T=-6.9°C$
