Question

A normal diet furnishes $200\, kcal$ to a $60\, kg$ person in a day. If this energy was used to heat the person with no losses to the surroundings, how much would the person's temperature increases? The specific heat of the human body is $0.83\, cal\,g ^{-1}{ }^{\circ} \,C ^{-1}$.

• A. $8.2^{\circ} C$
• B. $4.01^{\circ} C$
• C. $6.0^{\circ} C$
• D. $5.03^{\circ} C$

Answer 1

Answer: (B)

Given, $m=60\, kg =60 \times 10^{3} \,g , s=0.83\, cal \,g ^{-1}{ }^{\circ} \,C ^{-1}$
$Q=200 \,kcal =2 \times 10^{5} \,cal$
Amount of heat required for a person,
$Q=m s \Delta T$
$\Rightarrow \Delta T =\frac{Q}{m s}=\frac{2 \times 10^{5}}{60 \times 10^{3} \times 0.83}$
$=4.01^{\circ} C$
So, the person's temperature increases by $4.01^{\circ} C$.