The average specific heat of the human body is 3.6 kJ/kg·°C. If the body temperature of a (n) 96-kg man rises from 37°C to 39°C during strenuous exercise, determine the increase in the thermal energy of the body as a result of this rise in body temperature.

691200 J

Explanation:

From specific heat capacity,

ΔQ = cmΔt ... Equation 1

Where ΔQ = increase in thermal energy, c = specific heat capacity of the body, m = mass of the man, Δt = rise in temperature.

Given: c = 3.6 kJ/kg.°C = 3600 J/kg.°C, m = 96 kg, Δt = 39-37 = 2 °C.

Substitute into equation 1

ΔQ = 3600*96*2

ΔQ = 691200 J.

Hence the change in the thermal energy of the body = 691200 J