Question

The latent heat of vaporization of H₂O at body temperature (37°C) is 2.42 x 10⁶ J/kg. To cool the body of a 60.4-kg jogger [average specific heat capacity = 3500 J/(kg C°)] by 1.08 C°, how many kilograms of water in the form of sweat have to be evaporated?

Answer 1

0.094 kg

Step-by-step explanation:

Latent heat of vaporization of
`H_(2)O` at 37°C is
`2.42*10^(6)\text{ }(J)/(kg)`.

When the sweat on our body evaporates, it absorbs energy from our body to overcome it's Latent heat of vaporisation. Thus our body cools down when sweat evaporates.

So, Energy absorbed by sweat to evaporate = Energy lost by body

Specific heat capacity of human body =
`3500\text{ }\frac{J}{kg\text{ }C^(o)}`. Jogger weights 60.4 kg. Body temperature decreases by
`1.08\text{ }C^(o)`

Energy absorbed from body =
`mS\Delta T=3500*60.4* 1.08 =228312\text{ }J`

`228312\text{ }J=\text{Energy absorbed by sweat}=mC=m*2.42*10^(6)\\m=0.094\text{ }kg`

∴ 0.094 kg of sweat has evaporated from the body.