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A child running a temperature of 101^(@)F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweet 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 is approximately the same as that of water and latent heat of evaporation of water at that temperature is about 580cal.g^(-1).

Answer» <html><body><p></p>Solution :Here, fall in <a href="https://interviewquestions.tuteehub.com/tag/temp-1240937" style="font-weight:bold;" target="_blank" title="Click to know more about TEMP">TEMP</a>. <br/> = `DeltaT=101-98=3^(@)F=3xx5/(9)""^(@)C=5//3""^(@)C` <br/> Mass of child, m = 30 kg <br/> <a href="https://interviewquestions.tuteehub.com/tag/sp-1219706" style="font-weight:bold;" target="_blank" title="Click to know more about SP">SP</a>. heat of human body = sp. heat of water, <br/> `c=1000cal.kg^(-1)""^(@)C^(-1)` <br/> `therefore` Heat lost by the child, <br/> `DeltaQ=mcDeltaT=30xx1000xx5/3=50000"cals"` <br/> If m. be the mass of water evaporation in 20 min. <br/> then, `m.L=DeltaQorm.=(DeltaQ)/L=50000/580=86.2g` <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/average-13416" style="font-weight:bold;" target="_blank" title="Click to know more about AVERAGE">AVERAGE</a> rate of <a href="https://interviewquestions.tuteehub.com/tag/extra-447191" style="font-weight:bold;" target="_blank" title="Click to know more about EXTRA">EXTRA</a> evaporation <br/> = `86.2/20=4.31g"min"^(-1)`</body></html>


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