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Q. A person weighing $50\, kg$ takes in $1500\, kcal$ diet per day. If this energy were to be used in heating the body of person without any losses, then the rise in his temperature is (specific heat of human body $= 0.83 \,cal\,g^{-1}\, C^{-1}$)

Thermal Properties of Matter

Solution:

Here, $m=50\,kg=50 \times 10^{3}\,g$
$\Delta Q=1500\,kcal=1.5 \times 10^{6}\,cal$
$s=0.83\,cal\,g^{-1}\,{}^{\circ}C^{-1}$
As $\Delta Q=ms\Delta T$
$\Delta T=\frac{\Delta Q}{ms}=\frac{1.5 \times 10^{6}}{50 \times 10^{3} \times 0.83}$
$=36.14^{\circ}C$