A child running a temperature if 101^(@)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^(@)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 water, and latent heat of evaporation of water at that temperature is about "580 cal g"^(-1).

A child running a temperature if 101^(@)F is given an antipyrine (i.e.

1.	A child running a temperature if 101^(@)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^(@)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 water, and latent heat of evaporation of water at that temperature is about "580 cal g"^(-1).
Answer» SOLUTION :Here mass of BOY m = 30 kg Decrease in fever (temperature) of boy `Deltatheta=101^(@)F-98^(@)F` `=3^(@)F` `:.Deltatheta=(3xx5)/(9)=(5^(@))/(3)C` Specify heat of HUMAN body `c=4.2xx10^(3)" J kg"^(-1)" "^(@)C^(-1)` Latent heat of vaporization `L=580" cal g"^(-1)` `=580xx4.2xx10" J kg"` Suppose mass of sweating in 20 minutes by vaporization from body of boy = m. Time taken by sweet to convert into steam heat required to vaporize the sweat of mass m. `Q=m.L=m.xx580xx4.2xx10^(3)m^(-1)J . . . (1)` Heat lost by boy due to decrease in fever `Q.="mc "Deltatheta` `:.Q.=30xx4.2xx10^(3)xx5/3` `=2.1xx10^(5)J`. . . (2) Required heat for vaporization Q = heat lost by boy Q. `:.m.xx580xx4.2xx10^(3)=2.1xx10^(5)` `:.m.=(2.1xx10^(5))/(580xx4.2xx10^(3))` `:.m.=0.0862` kg `:.` Rate of vaporization of sweating `=(m.)/(t)` `=(0.0862)/(20)` `=0.0431" kg minute"^(-1)` `=4.31" g minute"^(-1)`