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

Answer:

90.079 g of water will evaporate from the body.

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

Answer 2

Heat required to decrease the temperature of body is given as

$Q = ms\Delta T$

here given that

$m = 100 kg$

$s = 4 J/g ^0C = 4000 J/Kg ^0C$

$\Delta T = 0.55^0C$

now by above equation

$Q = 100 \times 4000 \times 0.55$

$Q = 220,000 J$

now in order to evaporate water the heat is given as

$Q = mL$

$220000 = m \times 2.25 \times 10^6$

$m = 0.097 kg$

$m = 97 g$

so 97 g of water will evaporate from the body