Answer
(a) $0.103J$
(b) $0.480L$
Work Step by Step
(a) We can determine the required heat as
$m=\rho V$
$\implies m=1.29(0.003)=0.00387Kg$
$Q=cm\Delta T$
We plug in the known values to obtian:
$Q=(0.718)(0.00387)(37)=0.103J$
(b) We can determine the required volume as
$V_2=\frac{T_2}{T_1}V_1$
We plug in the known values to obtain:
$V_2=\frac{273+37}{273}$
$\implies V_2=1.136V_1$
Thus, the volume increases by $13.6\%$ that is $0.136$ and $3L=0.480L$
