Asked Mar 27, 2019

The latent heat of vaporization of H2O at body temperature (37.0 °C) is 2.44E+6 J/kg. To cool the body of a 71.7 kg jogger [average specific heat capacity = 3480 J/(kg·°C)] by 1.90 °C, how many kilograms of water in the form of sweat have to be evaporated?


Expert Answer

Step 1


Latent heat of vaporization, hv = 2.44 x 106 J/kg

Heat capacity of jogger, cv = 3480 J/(kg oC)

Change in temperature, Δt = 1.90o C

Mass of jogger’s body, M = 71.7 kg

Mass of vaporize water = m

Step 2

Heat loss in vaporization = Change in i...


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