Answer :
Answer:
2,296.7 grams of water can be cooled from 35°C to 20 °C by the evaporation of 60 g of water.
Explanation:
The heat of evaporation of water = [tex]H_e=2.4 kJ/g[/tex]
Mass of water evaporated , m= 60 g
Amount of heat required to evaporate 60 grams of water :Q
[tex]Q = H_e\times m=2.4 kJ/mol\times 60 =144 kJ[/tex]
Amount of heat lost to condense water 35 to 20 °C = -Q = -144 kJ = 144,000 J
Mass of water = m
Specific hat of water = c = 4.18 J/g K
Initial temperature = [tex]T_1=35^oC=308.15 K[/tex]
Final temperature = [tex]T_2=20^oC=293.15 K[/tex]
Change in temperature = [tex]\Delta T=T_2-T_2[/tex]
[tex]Q'=mc\Delta T[/tex]
[tex]m=\frac{Q'}{c\Delta T}=\frac{-144,000 J}{4.18 J/g K\times (293.15 K-308.15 K)}[/tex]
m = 2,296.7 g
2,296.7 grams of water can be cooled from 35°C to 20 °C by the evaporation of 60 g of water.