THermodynamics question Metabolic activity in the human body

THermodynamics question.

Metabolic activity in the human body releases approximately 1.0 times 10^4 kJ of heat per day. Assuming the body is 50 kg of water, how fast would the body temperature rise if it were an isolated system? How much water must the body eliminate as perspiration to maintain the normal body temperature (98.6 degree F)? Comment on your results. Assume the molar heat capacity of liquid water is 75.3 J/molK and the heat of vaporization of water is 2.41 kJ/g.

Solution

Here ,

heat capacity of water , S = 4186 J/kg.degree C

mass of water , m = 50 Kg

heat = 1 *10^4 kJ/day

change in body temperature= heat/(mass * heat capacity)

change in body temperature= 1 *10^4 * 10^3/(4186 * 50)

change in body temperature= 47.8 degreeC/day

the change in body temperature is 47.8 degreeC/day

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let the amount of water released is m

m * (2410 * 10^3) = 1 *10^4 * 10^3

m = 4.15 Kg

the amount of water to be prespiration is 4.15 Kg