Asked Apr 24, 2019

The latent heat of vaporization of H2O at body temperature (37.0 °C) is 2.35E+6 J/kg. To cool the body of a 79.1 kg jogger [average specific heat capacity = 3470 J/(kg·°C)] by 2.00 °C, how many kilograms of water in the form of sweat have to be evaporated?


Expert Answer

Step 1

Given : mass, m = 79.1 kg

Temperature, T = 2 degree celcius

Specific heat capacity, c = 3470 J/(kg. oC)

Latent heat of vaporization, L = 2.35 × 106 J/kg

Find: how many kilograms of water in the form of sweat have to be evaporated

Step 2

According to the given problem, The heat lost by jogger is equal to the heat gained by water:


Step 3

Substitute all the va...


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