A child running a temperature of $101\,^{\circ} F$ is given an antipyrin (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 $20$ minutes, what is the average rate of extra evaporation caused, by the drug (in $g/min$). Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is $30\; kg$. The spectfic heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about $580\; cal \;g^{-1}$.

Initial temperature of the body of the child, $T_{1}=101^{\circ} F$

Final temperature of the body of the child, $T_{2}=98^{\circ} F$

Change in temperature, $\Delta T=\left[(101-98) \times \frac{5}{9}\right]_{^o C }$

Time taken to reduce the temperature, $t=20$ min

Mass of the child, $m=30 kg =30 \times 10^{3} g$

Specific heat of the human body $=$ Specific heat of water $=c$

$=1000 cal / kg /^{\circ} C$

Latent heat of evaporation of water, $L=580 cal g ^{-1}$

The heat lost by the child is given as

$\Delta \theta=m c \Delta T$

$=30 \times 1000 \times(101-98) \times \frac{5}{9}$

$=50000 cal$

Let $m_{1}$ be the mass of the water evaporated from the child's body in 20 min.

Loss of heat through water is given by:

$\Delta \theta=m_{1} L$

$\therefore m_{1}=\frac{\Delta \theta}{L}$

$=\frac{50000}{580}=86.2 g$

$\therefore$ Average rate of extra evaporation caused by the drug $=\frac{m_{1}}{t}$ $=\frac{86.2}{200}=4.3\, g / min$