Answer:
The rate of transfer of the thermal energy is 616.67 W
Solution:
As per the question:
Temperature of the person after exercise, [tex]T = 40.0^{\circ}C[/tex]
After the temperature is reduced, [tex]T' = 37.0^{\circ}C[/tex]
Time taken, t = 30 min = 1800 s
Mass, m = 80.0 kg
Now,
The rate of transfer of thermal energy by the person can be calculated as:
Heat energy, Q = [tex]mC\Delta T = 80.0\times 3500\times (40.0 - 37.0) = 840\ kJ[/tex]
The rate of transfer of energy gives the power:
P = [tex]\frac{Q}{t} = \frac{840000}{1800} = 466.67\ W[/tex]
Total Power, P' = P + 150 = 466.67 + 150 = 616.67 W
