The latent heat of vaporization of H2O at body temperature (37.0 °C) is 2.37E+6 J/kg. To...

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The latent heat of vaporization of H2O at body temperature (37.0 °C) is 2.37E+6 J/kg. To...

The latent heat of vaporization of H₂O at body temperature (37.0 °C) is 2.37E+6 J/kg. To cool the body of a 77.9 kg jogger [average specific heat capacity = 3450 J/(kg·°C)] by 1.70 °C, how many kilograms of water in the form of sweat have to be evaporated?

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Answer 1

Answer #1

Conservation of energy:

Energy lost by jogger = deltaT * specific heat capacity * mjogger = Energy gained by evaporating water = latent heat * msweat

Solve for the mass of sweat:
msweat = (deltaT * specific heat capacity * mjogger) / latent heat

$m_{sweat}=\frac{1.7(3450)77.9}{2.37X10^{6}}\rightarrow \boldsymbol{m_{sweat}=0.1928Kg}$

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Answer 2

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