A body of specific heat 0.2 kcal / kg^ C is heated through 100^ C ....

Heat taken by the given body $=m c \Delta \theta$ {specific heat of body $=0.2 \mathrm{kcal} / \mathrm{kg}^{\circ} \mathrm{C}$, final temperature of the body \left.=100^{\circ} \mathrm{C}\right\} $$\begin{aligned} \Delta E & =m \times 0.2 \times 100=20 \mathrm{mkcal} \\ & =20 \mathrm{~m} \times 4.2 \times 10^3 \mathrm{~J}\end{aligned}$$$Now gain in mass is given by$\Delta m=\frac{\Delta E}{C^2} \left(\because E=m c^2\right)$=\frac{20 \mathrm{~m} \times 4.2 \times 10^3}{\left(3 \times 10^8\right)^2}$ Therefore, percentage increase in mass is given by $$\begin{aligned} & =\frac{\Delta m}{m} \times 100 \\ & =\frac{20 m \times 4.2 \times 10^3}{\left(3 \times 10^8\right)^2 \times m} \times 100 \\ & =9.3 \times 10^{-11} \%\end{aligned}$$