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Thermal Properties of Matter

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Question : 18 of 22
Marks: +1, -0
A child running a temperature of 101°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°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 evaporation of water at that temperature is about 580 cal g−1580\ \mathrm{cal}\ \mathrm{g}^{-1}.
Solution:  
Here, fall in temperature =ΔT=101−98=\Delta T=101-98
=3×5∘9 C=5∘3 C=3\times\frac{5^{\circ}}{9}\ \mathrm{C}= \frac{5^{\circ}}{3}\ \mathrm{C}
Mass of child, m=30m = 30 kg
Sp. heat of human body = sp. heat of water,
c=1000c=1000 cal. kg−1 ∘C−1\mathrm{kg}^{-1}\,{}^{\circ}\mathrm{C}^{-1}
∴\therefore Heat lost by the child =ΔQ=mcΔT=\Delta Q=m c \Delta T
=30×1000×53=30 \times 1000 \times \frac{5}{3}
=50000=50000 cals
If m′m' be the mass of water evaporated in 20 min., then m′L=ΔQm' L = \Delta Q
or 50000580=86.2 g\frac{50000}{580}=86.2\ \mathrm{g}
∴\therefore Average rate of extra evaporation =86.220=4.31 g min−1= \frac{86.2}{20}=4.31\ \mathrm{g}\ \mathrm{min}^{-1}
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