Question

Following vigorous exercise, the bodytemperature ofan BOD-kg person is 400°C . At what rate in watts must the persontransfer thermal energy to reduce the the body temperature to 370°C in 30.0 min, assuming the body continues toproduce energy at the rate of 150 W? [1 watt = 1joule/second or 1 W = 1 J/s).

Answer 1

The rate of transfer of the thermal energy is 616.67 W

Solution:

As per the question:

Temperature of the person after exercise,
`T = 40.0^(\circ)C`

After the temperature is reduced,
`T' = 37.0^(\circ)C`

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 =
`mC\Delta T = 80.0* 3500* (40.0 - 37.0) = 840\ kJ`

The rate of transfer of energy gives the power:

P =
`(Q)/(t) = (840000)/(1800) = 466.67\ W`

Total Power, P' = P + 150 = 466.67 + 150 = 616.67 W