A child running a temperature of `101^F` is given and antipyrin (i.e. a madicine that lowers fever) which cause 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 averatge rate of extra evaporation caused, by the drug ? Assume the evaporation mechanism to 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).`

Explanation

Initial temperature of the body of the child, `T_(1) = 101^(@)F`
Final temperature of the body of the child, `T_(2) = 98^(@)F`
Change in temperature, `DeltaT=[(101-98)xx(5)/(9)]^(@)C`
Time taken to reduce the temperature, t = 20 min
Mass of the child, m = 30 `kg = 30 × 10^(3) g`
Specific heat of the human body = Specific heat of water = c
`=1000 cal//kg// ^(@)C`
Latent heat of evaporation of water, `L = 580 cal g^(–1)`
The heat lost by the child is given as:
`Delta theta=mcDeltaT`
`=30xx1000xx(101-98)xx(5)/(9)`
=50000 cal
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:
`Delta theta=m_(1)L`
`:. m_(1)=(Delta theta)/(L)`
`=(50000)/(580)=86.2 g`
`:. " Average rate of extra evaporation caused by the drug " =(m_(1))/(t)`
`=(86.2)/(200)=4.3g//min`