Question

A person weighing 50 kg takes in 1500 k Cal 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

• A. 30 °C
• B. 48 °C
• C. 40.16 °C
• D. 36.14 °C

Answer 1

Answer: (D)

36.14 °C

Answer 2

Here , m = 50 kg = 50 × $10^{3}g$

$\Delta Q = 1500kcal = 1.5 \times 10^{6}cal$

$s = 0.83calg^{- 1}{^\circ}C^{- 1}$

As $\Delta Q = ms\Delta T$

$\Delta T = \frac{\Delta Q}{ms} = \frac{1.5 \times 10^{6}}{50 \times 10^{3} \times 0.83} = 36.14{^\circ}C$