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Exercises · 10.16

Q.A child running a temperature of 101 ∘F101\ ^\circ\text{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 ∘F98\ ^\circ\text{F} in 2020 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 30 kg30\ \text{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\ \text{cal g}^{-1}.

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The heat the child's body loses as its temperature falls from 101∘101^\circF to 98∘98^\circF is carried away entirely by extra sweat evaporation. Working through the numbers gives an average extra evaporation rate of about 4.31 g min−14.31\ \text{g min}^{-1}.

Step 1 - Convert the temperature drop to Celsius

The body's specific heat is given in cal g−1 ∘C−1\text{cal g}^{-1}\,^\circ\text{C}^{-1}, so the Fahrenheit change must be converted using ΔTC=59ΔTF\Delta T_C = \frac{5}{9}\Delta T_F:

ΔTF=101−98=3 ∘F⇒ΔTC=3×59=53 ∘C.\Delta T_F = 101-98 = 3\,^\circ\text{F} \quad\Rightarrow\quad \Delta T_C = 3\times\frac{5}{9} = \frac{5}{3}\,^\circ\text{C}.

Step 2 - Heat lost by the child's body

m=30 kg=30,000 g,c=1 cal g−1 ∘C−1.m = 30\ \text{kg} = 30{,}000\ \text{g}, \qquad c = 1\ \text{cal g}^{-1}\,^\circ\text{C}^{-1}.

Q=mcΔT=30,000×1×53=50,000 cal.Q = mc\Delta T = 30{,}000\times1\times\frac{5}{3} = 50{,}000\ \text{cal}.

Step 3 - Mass of sweat this heat can evaporate

All of this heat is assumed carried away by evaporation, at latent heat L=580 cal g−1L=580\ \text{cal g}^{-1}: …

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