Question

A person weighing $650kg$ taken 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 $=0.83cal{g}^{-1}{C}^{-1})$:

• A. ${30}^{o}C$
• B. ${48}^{o}C$
• C. ${40.16}^{o}C$
• D. ${36.14}^{o}C$

Answer 1

Answer: (D)

${36.14}^{o}C$

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