Question

A child running a temperature of \[101\,^\circ {\text{F}}\] is given an antipyretic (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to \[98\,^\circ {\text{F}}\] in \[20\] minutes, what is the average rate of extra evaporation caused by the drug? Assume the evaporation mechanism to be the only way by which the heat is lost. The mass of the child is \[30\,{\text{kg}}\] . The specific heat of the human body is approximately the same as that of water, latent heat of water at that temperature is \[580\,{\text{cal}}\,{{\text{g}}^{ - 1}}\] .
(A) \[2.1\,{\text{g}}{\min ^{ - 1}}\]
(B) \[4.3\,{\text{g}}{\min ^{ - 1}}\]
(C) \[6.2\,{\text{g}}{\min ^{ - 1}}\]
(D) \[8.4\,{\text{g}}{\min ^{ - 1}}\]

Answer 1

Hint:First of all, we will find the change in temperature and convert the same into degree centigrade. After that we will find the heat lost by the body and we will also find the heat lost through water to calculate the mass of the water. Lastly, we will find the rate of mass evaporated per unit time.

Complete step by step solution:
In the given question, we are supplied with the following data:
The initial temperature of the body of the child is \[101\,^\circ {\text{F}}\] .
The final temperature of the body of the child is \[98\,^\circ {\text{F}}\] .
The time required to cool down the body from the higher temperature to the lower temperature is \[20\] minutes.
The mass of the child is given as \[30\,{\text{kg}}\] .
The latent heat of water at that temperature is \[580\,{\text{cal}}\,{{\text{g}}^{ - 1}}\] .
We are asked to find the average rate of extra evaporation caused by the drug.
To begin with, we will first find the change in temperature. After that we will proceed to find the heat energy lost.
Let us proceed to solve the problem.
Before that we will keep all the quantities already beforehand.
The mass of the child is given as \[30\,{\text{kg}}\] .
We will convert the unit into CGS units.
$30\,{\text{kg}} \\
\Rightarrow 3{\text{0}} \times {\text{1}}{{\text{0}}^{\text{3}}}\,{\text{g}} \\$
We will also calculate the change in temperature which is shown below:
$\Delta T = \left( {101 - 98} \right)\,^\circ {\text{F}} \\
\Rightarrow \Delta T = 3\,^\circ {\text{F}} \\
\Rightarrow \Delta T = 3 \times \dfrac{5}{9}\,^\circ {\text{C}} \\
\Rightarrow \Delta T = \dfrac{5}{3}\,^\circ {\text{C}} \\$
We also know that the specific heat of the human body is equal to the specific heat of water whose value is\[1000\,{\text{cal}}\,{\text{k}}{{\text{g}}^{ - 1}}\,^\circ {{\text{C}}^{ - 1}}\] .
Now, we can find the heat lost by the child by using the formula given below:
\[\Delta \theta = mc\Delta T\] …… (1)
Where,
\[\Delta \theta \] indicates the heat lost.
\[m\] indicates the mass of the body.
\[c\] indicates the specific heat of the body.
\[\Delta T\] indicates the change in temperature.
Now, we will substitute the required values in the equation (1) and we get:
$\Delta \theta = mc\Delta T \\
\Rightarrow \Delta \theta = 30 \times {10^3} \times 1000 \times \dfrac{5}{3} \\
\Rightarrow \Delta \theta = 50000\,{\text{cal}} \\$
Let us assume that the mass of the water which has evaporated from the body of the child in \[20\] minutes be \[{m_1}\] .
We have a formula which gives the heat lost through water is given below:
\[\Delta \theta = {m_1}L\] …… (2)
Where,
\[\Delta \theta \] indicates the heat lost through water.
\[{m_1}\] indicates the mass of water lost.
\[L\] indicates the latent heat of evaporation of water.
Now, we substitute the required values in the equation (2) and we get:
$\Delta \theta = {m_1}L \\
\Rightarrow 50000 = {m_1} \times 580 \\
\Rightarrow {m_1} = \dfrac{{50000}}{{580}} \\
\Rightarrow {m_1} = 86.2\,{\text{g}} \\$
Now, we will find the average rate of evaporation caused by the drug as mentioned above.
\[r = \dfrac{{{m_1}}}{t}\] …… (3)
Where,
\[r\] indicates the rate.
\[{m_1}\] indicates the mass of water.
\[t\] indicates the time required in minutes.
Now, we substitute the required values in the equation (3) and we get:
$r = \dfrac{{{m_1}}}{t} \\
\Rightarrow r = \dfrac{{86.2\,{\text{g}}}}{{20\,\min }} \\
\therefore r = 4.3\,{\text{g}}\,{\min ^{ - 1}} \\$

Hence, the average rate of extra evaporation caused by the drug is \[4.3\,{\text{g}}\,{\min ^{ - 1}}\] .The correct option is (B).

Note:While solving the given problem, many students tend to make the mistake while converting the degree Fahrenheit to degree centigrade. The heat energy lost by the body to lower its temperature is equal to the heat lost by water. No heat is dissipated in other forms, do remember that.