Answer
$17\times 10^{-3}K$
Work Step by Step
We know that
$\Delta m=m_i-m_f$
$\implies \Delta m=(239.052158u)-(235.043925u+4.002603u)=0.00563u$
and $E=(0.00563u)(\frac{931.5\frac{MeV}{c^2}}{1u})c^2$
$\implies E=5.224MeV$
The total energy is given as
$E^{\prime}=nE$
$\implies E^{\prime}=(4.632\times 10^{14})(5.224MeV)=24.29\times 10^{14}MeV$
Now $\Delta T=\frac{E^{\prime}}{m_w c_w}$
We plug in the known values to obtain:
$\Delta T=\frac{(24.29\times 10^{14}MeV)(\frac{1\times 10^6eV}{1MeV})(\frac{1.6\times 10^{-19}J}{1eV})}{(4.75Kg)(4186J/Kg.K)}$
$\implies \Delta T=17\times 10^{-3}K$
