A man consumes 2245 kcal of food in one day, converting most of it to maintain body temperature. If he loses half this energy by evaporating water (through breathing and sweating), how many kilograms of water evaporate?

Answer :

Answer:

2.079 kg

Explanation:

Latent heat of vaporisation of water, L = 2,260 kJ/kg = 2260000 J/kg

Energy intake = 2245 Kcal

Energy used for evaporating water = half of energy intake

E = 2245 / 2 = 1122.5 kcal

now convert kcal into Joule

1 kcal = 4186 J

So, E = 1122.5 x 4186 = 4698785 J

let m be the mass of water which is evaporated.

by the use of formula of heat

E = m x L

Where, L be the latent heat of vaporisation.

4698785 = m x 2260000

m = 2.079 kg

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