A typical human (160. lbs) produces about 10. MJ of energy in the form of heat e

Question

A typical human (160. lbs) produces about 10. MJ of energy in the form of heat each day through metabolic activity. Part a) Assuming the human body is an adiabatic system, calculate the rise in temperature associated with metabolic activity during one day. To do this calculation, assume that human body has approximately the same molar heat capacity as liquid water (C = 75.29 J/mol.K). Part b) Let us assume now that the human body is an open system and that perspiration (evaporation of water produced by sweating) is the mechanism which ensures that the body temperature remains constant. Given that the heat required to vaporize one mole of water liquid is given by q = 43 kJ/mol(H2O), calculate the mass of water that a 160. lbs human body must lose every day to keep its temperature constant. Note that Parts a) and b) are independent from one another!

Explanation / Answer

part a)

let the rise in temperature is T

for the human body

mass , m = 160 lbs = 72.3 Kg

Now, E = m * C * T

10 *10^6 = 72.3 * 4186 * T

T = 33.04 degree

the increase in temperature each day will be 33.04 degree

b)

let the mass of water vaporized is m

E = n * H

10 *10^6 = m/(0.018) * 43 *10^3

m = 4.186 Kg

the mass of water vapourzied is 4.19 Kg

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