During illness an 80, kg man ran a fever of 102.2^o F instead of normal body temperature of 98.

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10-1.Thermometry, Thermal Expansion and Calorimetry

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

A

$100$

B

$160$

C

$50$

D

$92$

Solution

(b) Since $102.2^o F \to 39^o C$ and $98.6^o F \to 39°C$

Hence $\Delta Q = m. s.\Delta\,\theta = 80 × 1000 × (39 -37)$

$= 16 × 10^4 \,cal = 160\, kcal.$

Standard 11

Physics

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