Question

Estimate the mass of water that must evaporate from the skin to cool the body by 0.55 ∘c. assume a body mass of 100 kg and assume that the specific heat capacity of the body is 4.0 j/g⋅∘c.

Answer 1

90.079 g of water will evaporate from the body.

Step-by-step explanation:

Change in temperature = ∆T = 0.55 oC

Specific heat capacity of the body = s = 4 J/goC = 4000 J/kgoC

Mass of body = m = 100 kg

Now, heat lost by the body will be:

Q1 = - ms∆T ……….(i)

By putting values in equation (i)

Q1 = - (100)(4000)(0.55)

Q1 = - 220000 J = - 220 KJ

Heat absorbed by the water will be:

Q2 = - (Q1) = 220 KJ

Evaporation is an endothermic process and standard enthalpy per mole evaporation of water is 44.01 KJ so,

Mass of evaporated water = (220 KJ)(1 mol/44.01 KJ) (18.02g/1 mol)

Mass of evaporated water = 90.079 g