Question

A child running a temperature of $101^{\circ} F$ is given an antipyrine (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 $98^{\circ} F$ in $20\, \min$, what is the average rate of extra evaporation caused by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is $30\, kg$. The specific heat of human body is approximately the same as that of water, and latent heat of water at that temperature is about $580\, cal / g$.

• A. 4.3 g/min
• B. 6.2 g/min
• C. 3.6 g/min
• D. 2.4 g/min

Answer 1

Answer: (A)

Mass of the child, $M=30\, kg =30 \times 10^{3} g$
It is given specific heat of human body is equal to specific heat of water, hence specific heat of human body $c=1\, cal / g -{ }^{\circ} C$
Fall in temperature of the body,
$\Delta T=101-98=3^{\circ} F =3 \times \frac{5}{9}=\frac{5}{3}^{\circ} C$
Time taken for the lowering of the child's temperature,
$t=20\, \min$
Latent heat of evaporation of water, $L=580\, cal / g$
Let $m$ be the mass of the sweat that evaporates from the child's body in $20\, \min$.
Heat gained by the sweat $=m L\,\,(i)$
Heat lost by the child $=M c \Delta T\,\,(ii)$
From equations (i) and (ii), using principle of calorimetry
$m L=M c \Delta T$
or $m=\frac{M c \Delta T}{L}=\frac{\left(30 \times 10^{3}\right) \times 1 \times(5 / 3)}{580}$
$=86.2\, g$
Rate of evaporation
$=\frac{m}{\Delta t}=\frac{86.2}{20}=4.3\, g / \min$