Question

A child runing a temperature of $101{}^{\circ }F$ is given an antipyrin (i.e. a madicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to $98{}^{\circ }F\text{in}20\text{minutes},$ 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 $30kg.$ The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about $580cal/g$

Answer 1

Given , Initial temperature of the child, $T_i=101{}^{\circ }F$

Final temperature of the child,$T_f=98{}^{\circ }F$

Decrease in the tempareture, $\Delta T=(101-98)=$

$3{}^{\circ }F=3\times \left(\frac{5}{9}\right)=1.67{}^{\circ }C$

Mass of the child , $m=30kg$

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

Specific heat of the human body = Specific heat of

water $=c=1000\frac{cal}{kg{-}^{\circ }C}$

Step 1: Find heat lost by the child.

The heat lost by the child is given as

$\Delta Q=\text{mc}\Delta T$

$=30\times 1000\times 1.67$

$=50100\text{cal}$

Step 2: Find amount of water evaporated from the child's body.

Let ${}^{\prime }{m}^{\prime }$ be the amount of water evaporated from the child's body in $20\text{min}.$ Latent heat of evaporation of water, $L=580\text{cal/g}$

$m^{\prime }=\frac{\Delta Q}{L}$

$=\left(\frac{50100}{580}\right)=86.37g$

Step 3: Find average rate of evaporation.

Therefore, the average rate of evaporation $=\frac{86.2}{20}=4.3g\text{/}min$