Heat required to decrease the temperature of body is given as
[tex]Q = ms\Delta T[/tex]
here given that
[tex] m = 100 kg[/tex]
[tex]s = 4 J/g ^0C = 4000 J/Kg ^0C[/tex]
[tex]\Delta T = 0.55^0C[/tex]
now by above equation
[tex]Q = 100 \times 4000 \times 0.55[/tex]
[tex]Q = 220,000 J[/tex]
now in order to evaporate water the heat is given as
[tex]Q = mL[/tex]
[tex]220000 = m \times 2.25 \times 10^6[/tex]
[tex]m = 0.097 kg[/tex]
[tex]m = 97 g[/tex]
so 97 g of water will evaporate from the body
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
