question_answer 18)

A child running a temperature of \[{{101}^{o}}F\] is given an antipyrin (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 to \[{{98}^{o}}F\] in 20 min., what is the average rate of extra evaporation caused, by the drug? Assume evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. 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. \[m{{m}^{2}}\]

Answer:

Here, fall in temp. \[{{\text{Y}}_{\text{steel}}}\text{=2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{11}}}\text{N}{{\text{m}}^{\text{-2}}}\text{ and}\]\[{{Y}_{alu\min ium}}=7\cdot 0\times \text{1}{{\text{0}}^{\text{10}}}\text{ N}{{\text{m}}^{\text{-2}}}\] Mass of child, m = 30 kg Sp. heat of human body = sp. heat of water, \[{{\text{l}}_{\text{1}}}\text{=l;}{{\text{A}}_{\text{1}}}\text{=m}{{\text{m}}^{\text{2}}}\text{;}\]\[{{\text{Y}}_{\text{1}}}\text{=2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{11}}}\text{N}{{\text{m}}^{\text{-2}}}\] \[\text{B,}{{\text{l}}_{\text{2}}}\text{=l;}{{\text{A}}_{\text{2}}}\text{=m}{{\text{m}}^{\text{2}}}\text{;}\]Heat lost by the child, \[{{\text{Y}}_{\text{2}}}\text{=7 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{10}}}\text{N}{{\text{m}}^{\text{-2}}}\] If m' be the mass of water evaporated in 20 min. then, \[{{\text{F}}_{\text{1}}}\text{ and }{{\text{F}}_{\text{2}}}\] or \[\frac{{{\text{F}}_{\text{1}}}}{{{\text{A}}_{\text{1}}}}\text{=}\frac{{{\text{F}}_{\text{2}}}}{{{\text{A}}_{\text{2}}}}\] \[\frac{{{\text{F}}_{\text{1}}}}{{{\text{F}}_{\text{2}}}}\text{=}\frac{{{\text{A}}_{\text{1}}}}{{{\text{A}}_{\text{2}}}}\text{=}\frac{\text{1}}{\text{2}}\] Average rate of extra evaporation \[{{F}_{1}}x={{F}_{2}}\left( 1\cdot 05-x \right)\]