Question

Asked Mar 27, 2019

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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?

Step 1

Given,

Latent heat of vaporization, h_{v }= 2.44 x 10^{6} J/kg

Heat capacity of jogger, c_{v} = 3480 J/(kg ^{o}C)

Change in temperature, Δt = 1.90^{o} 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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