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).

The rate of transfer of the thermal energy is 616.67 W

Solution:

As per the question:

Temperature of the person after exercise,

After the temperature is reduced,

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 =

The rate of transfer of energy gives the power:

P =

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

  the unit of the measure =

explanation:

  the property that is quantified by the energy needed per of material to raise the temperature by one.

  si unit of energy = joule (j)

  si unit of temperature = kelvin (k)

  si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is

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