A child running a temperature of `101 ^(@) F` is given an antipyrin (1.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 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)`.

Explanation

Here, fall in temp.
`= Delta T = 101 - 98 = 3^(0) F = 3 xx (5)/(9)""^(0) C = 5 // 3^(0) C `
Mass of child, m = 30 kg
Sp. Heat of human body = sp. Heat of water,
c = 1000 cal.`kg^(-1) ""^(0)C^(-1)`
`therefore ` Heat lost by the child ,
`Delta Q = mc Delta T = 30 xx 1000 xx (5)/(3) = 50000` cals
If m . be the mass of water evaporated in 20 min.
then , m.L = `Delta Q ` or m. = `(Delta Q)/(L) = (50000)/(580)` = 86.2g
`therefore ` Average rate of extra evaporation
`= (86.2)/(20) = 4.31" g min"^(-1)`