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

Here fall in temp. ` = Delta T =101 - 98 =3^@F = 3xx(5)/(9) .^@C = 5//3 ^@C`
Mass of child, m 30 kg Sp. Heat of human body = sp. Heat of water, `s = 1000 cal. Kg^(-1) .^@C^(-1)`
`:.` Heat lost by the child, `Delta Q = ms Delta T = 30xx 1000xx(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) = 86.2 g`
`:.` Average rate of exra evaporation ` =(86.2)/(20) = 4.31 g mi n^(-1)`