Question

A person weighing 50kg takes in $1500\,{\text{kcal}}$ 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$= 0.83\,{\text{cal}} \cdot {{\text{g}}^{ - 1}} \cdot {{\text{C}}^{ - 1}}$)
A. 30
B. 48
C. 40.16
D. 36.14

Answer 1

Use the formula for the energy exchanged by the substance with the surrounding. This formula gives the relation between heat energy exchanged, mass of the substance. Specific heat of the substance and change in temperature of the substance. Convert the unit mass into grams and energy into calories and substitute these values in the formula and calculate rise in temperature of the body.

Formula used:
The heat energy $Q$ exchanged by the substance with its surrounding is given by
$Q = mc\Delta T$ …… (1)
Here, $m$ is the mass of the substance, $c$ is specific heat of the substance and $\Delta T$ is a change in temperature of the substance.

Complete step by step answer:
We have given that the mass of the person is $50\,{\text{kg}}$ and the energy taken in by the person through his diet is $1500\,{\text{kcal}}$.
$m = 50\,{\text{kg}}$
$\Rightarrow Q = 1500\,{\text{kcal}}$
The specific heat of the human body is $0.83\,{\text{cal}} \cdot {{\text{g}}^{ - 1}} \cdot {{\text{C}}^{ - 1}}$.
$c = 0.83\,{\text{cal}} \cdot {{\text{g}}^{ - 1}} \cdot {{\text{C}}^{ - 1}}$
We have asked to calculate the rise in temperature of the person when the amount of energy taken by him through his diet is completely used to increase the temperature of his body.

Let us first convert the temperature of the energy taken in by the person from kilocalorie to calorie.
$Q = \left( {1500\,{\text{kcal}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{cal}}}}{{1\,{\text{kcal}}}}} \right)$
$\Rightarrow Q = 1.5 \times {10^6}\,{\text{cal}}$
Hence, the energy taken in by the person is $1.5 \times {10^6}\,{\text{cal}}$.
Convert the unit of mass of the person from kilogram to gram.
$m = \left( {50\,{\text{kg}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{g}}}}{{1\,{\text{kg}}}}} \right)$
$\Rightarrow m = 5 \times {10^4}\,{\text{g}}$
Hence, the mass of the person is $5 \times {10^4}\,{\text{g}}$.

Now we can calculate the rise in temperature of the person using equation (1).Rearrange equation (1) for the change in temperature.
$\Delta T = \dfrac{Q}{{mc}}$
Substitute $1.5 \times {10^6}\,{\text{cal}}$ for $Q$, $5 \times {10^4}\,{\text{g}}$ for $m$ and $0.83\,{\text{cal}} \cdot {{\text{g}}^{ - 1}} \cdot {{\text{C}}^{ - 1}}$ for $c$ in the above equation.
$\Delta T = \dfrac{{1.5 \times {{10}^6}\,{\text{cal}}}}{{\left( {5 \times {{10}^4}\,{\text{g}}} \right)\left( {0.83\,{\text{cal}} \cdot {{\text{g}}^{ - 1}} \cdot {{\text{C}}^{ - 1}}} \right)}}$
$\therefore \Delta T = 36.14^\circ {\text{C}}$
Therefore, the rise in temperature of the person’s body is $36.14^\circ {\text{C}}$.

Hence, the correct option is D.

Note: The students should not forget to convert the unit of energy taken in by the person from kilocalorie to calorie and mass of the person from kilogram to gram as the unit of the specific heat of the human body includes the units calorie and gram. If these conversions are not done then the final answer for rise in temperature will be incorrect.