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(B) First,convert the temperatures from Fahrenheit to Celsius using the formula $C = (F - 32) 	imes \frac{5}{9}$.Normal temperature: $98.6^\circ F =

Vedclass · Accepted Answer

$160$

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(B) First,convert the temperatures from Fahrenheit to Celsius using the formula $C = (F - 32) 	imes \frac{5}{9}$.Normal temperature: $98.6^\circ F = (98.6 - 32) 	imes \frac{5}{9} = 66.6 	imes \frac{5}{9} = 37^\circ C$.Fever temperature: $102.2^\circ F = (102.2 - 32) 	imes \frac{5}{9} = 70.2 	imes \frac{5}{9} = 39^\circ C$.The change in temperature is $\Delta 	heta = 39^\circ C - 37^\circ C = 2^\circ C$.The mass of the man is $m = 80 \, kg = 80,000 \, g$.The specific heat capacity of water is $s = 1 \, cal/g^\circ C$.The heat required is given by $\Delta Q = m \cdot s \cdot \Delta 	heta$.$\Delta Q = 80,000 \, g 	imes 1 \, cal/g^\circ C 	imes 2^\circ C = 160,000 \, cal$.Since $1,000 \, cal = 1 \, kcal$,we have $\Delta Q = 160 \, kcal$.

During illness,an $80 \, kg$ man had a fever of $102.2^\circ F$ instead of his normal body temperature of $98.6^\circ F$. Assuming that the human body is mostly water,how much heat is required to raise his temperature by that amount in $kcal$?

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