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  • A child running a temperature of 101 degree 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.

Q. 11.16 A child running a temperature of 101^{\circ}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^{\circ}F  in 20 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\; 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^{-1}.

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Initial Temperature of the boy = 101 oF

Final Temperature of the boy = 98 oF

Change in Temperature is

\\\Delta T=3\ ^{o}F\\ \Delta T=3\times \frac{5}{9}\\ \Delta T=1.67\ ^{o}C

 Mass of the child is m = 30 kg

Specific heat of human body = 1000 cal kg-1 oC-1

Heat released is \\\Delta Q

\\\Delta Q=mc\Delta T\\\Delta Q=30\times 1000\times 1.67\\ \Delta Q=50000\ cal

Latent heat of evaporation of water = 580 cal g-1

The amount of heat lost by the body of the boy has been absorbed by water.

Let the mass of water which has evaporated be m'

\\\Delta Q=m'L\\ m'=\frac{Q}{L}\\ m'=\frac{50000}{580}\\ m'=86.2\ g
Time in which the water has evaporated, t = 20 min.

Rate of evaporation is m'/t

\\\frac{m'}{t}=\frac{86.2}{20}\\ \frac{m'}{t}=4.31\ g\ min^{-1}

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