A child running a temperature of $101^{0} F$ is given an antipyrin (i.e. a medicine that lowers fever ) which causes an increases in the rate of evaporation of sweat from his body. If the fever is brought down to $98^{o} F$ 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 $580 c a l g^{- 1}$ .

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Solution

Initial temperature of the body of the child, $T_{1} = 101^{o} F$
Final temperature of the body of the child, $T_{2} = 98^{o} F$
Change in temperature, $△ T = [(101 - 98) \times 5 / 9 = 5 / 3^{o} C$
Time taken to reduce the temperature, t = 20 min
Mass of the child, $m = 30 k g = 30 \times 10^{3} g$
Specific heat of the human body = Specific heat of water = c
$= 1000 c a l / k g /^{o} C$
Latent heat of evaporation of water, $L = 580 c a l g^{- 1}$
The heat lost by the child is given as:
$△ θ = m c △ T$
$= 30 \times 1000 \times (101 - 98) \times (5 / 9)$
$= 50000 c a l$
Let $m_{1}$ be the mass of the water evaporated from the child’s body in 20 min.
Loss of heat through water is given by:
$△ θ = m_{1} L$
$∴ m_{1} = △ θ / L$
$= (50000 / 580) = 86.2 g$
$∴$ Average rate of extra evaporation caused by the drug $= m_{1} / t$
$= 86.2 / 20$
$= 4.3 g / m i n .$

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