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A child running a temperature of $101^{\circ} \mathrm{F}$ 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 $98^{\circ} \mathrm{F}$ 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 in $30 \mathrm{~kg}$. 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 580 cal $\mathrm{g}^{-1}$.
PhysicsThermal Properties of Matter
Solution:
1858 Upvotes Verified Answer
Decrease in temperature $=101-98=3^{\circ} \mathrm{F}$ $=3 \times \frac{5}{9}=\frac{5}{3}{ }^{\circ} \mathrm{C} ;$ Mass of child $=\mathrm{m}=30 \mathrm{~kg}$ Specific heat of water $=$ specific heat of human body $=$ $\mathrm{c}=1000 \mathrm{cal} / \mathrm{kg} /{ }^{\circ} \mathrm{C}$
$\therefore \quad$ Heat lost by the child $=\mathrm{mc} \Delta t=30 \times 1000 \times \frac{5}{3}=5000 \mathrm{cal}$.
Let $\mathrm{m}^{\prime}$ be the mass of water evaporated then $\mathrm{m}^{\prime} \mathrm{L}=$ $\mathrm{mc \Delta t}$ or $\mathrm{m}^{\prime}=\frac{5000}{580}=8.620$
$\therefore \quad$ Average rate of extra evaporation $=\frac{8.62}{20}=0.431 \mathrm{~g} / \mathrm{min}$.

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