Question

A man consumes 2985 kcal of food in one day, converting most of it to body temperature. If he loses half this energy by evaporating water (through breathing and sweating), how many kilograms of water evaporate?

Answer 1

Answer:

m = 2.76 kg

Explanation:

The mass of water evaporated can be found by using the following formula:

$Q = mH\\\\m = \frac{Q}{H}$

where,

m = mass of water evaporated = ?

Q = Energy used = $\frac{1}{2}(2985\ kcal) = (1492.5\ kcal)(\frac{4.184\ KJ}{1\ kcal})$ = 6244.62 KJ

H = Latent heat of vaporization of water = 2260 KJ/kg

Therefore,

$m = \frac{6244.62\ KJ}{2260\ KJ\kg}$

m = 2.76 kg