Answer
$T_{sat}=-11.7^{\circ}C$
$Q_{abs}=1089.92kJ$
Work Step by Step
From table A-3 the properties of isobutane at $1atm$
$T_{sat}=-11.7^{\circ}C$
$\rho=593.8\frac{kg}{m^3}$
$h_{fg}=367.1\frac{kJ}{kg}$
$m=\rho V=(593.8\frac{kg}{m^3})(0.005m^3)=2.969kg$
$Q_{abs}=mh_{fg}=2.969kg*367.1\frac{kJ}{kg}=1089.92kJ$
