Question

The latent heat of vaporization of H2O at body temperature (37.0° C) is 2.42 106 J/kg. To cool the body of a 84 kg jogger (average specific heat capacity = 3500 J/(kg · C°)) by 1.2° C, how many kilograms of water in the form of sweat have to be evaporated?

Answer 1

Answer:

$m = 0.146 kg$

Explanation:

As we know that temperature of the jogger is decreased then the amount of heat to be removed from the body

The equation is given as

$Q = ms\Delta T$

$Q = 84 \times 3500 \times 1.2$

$Q = 352800 J$

now let say m mass of the sweat is evaporated out

so by energy conservation we have

$Q = mL$

$352800 = m(2.42 \times 10^6)$

$m = 0.146 kg$