Answer
$T_{sat}=-42.1^{\circ}C$
$Q_{abs}=1242.76kJ$
Work Step by Step
From table A-3 the properties of propane at $1atm$
$T_{sat}=-42.1^{\circ}C$
$\rho=581\frac{kg}{m^3}$
$h_{fg}=427.8\frac{kJ}{kg}$
$m=\rho V=(581\frac{kg}{m^3})(0.005m^3)=2.905kg$
$Q_{abs}=mh_{fg}=2.905kg*427.8\frac{kJ}{kg}=1242.76kJ$
