Q. During illness an $80\ kg$ man ran 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
A) $100\ kcal$
B) $160\ kcal$
C) $50\ kcal$
D) $92\ kcal$
Explanation
(b) Since ${102.2}^{o}F \rightarrow 39℃\ $and$\ {98.6}^{o}F \rightarrow 37℃$
Hence $\mathrm{\Delta}Q = m.s.\mathrm{\Delta}\theta = 80 \times 1000 \times (39 - 37)$
$= 16 \times 10^{4}cal = 160\ kcal$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.