Question

or many years drinking water has been cooled in hot cli- mates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from 35 to 20 °C by the evaporation of 60 g of water?the heat of evaporation of water in this temperature rsnges is 2.4 kj/g. the specific heat capacity of water is 4.18 j/g K.

Answer 1

2,296.7 grams of water can be cooled from 35°C to 20 °C by the evaporation of 60 g of water.

Step-by-step explanation:

The heat of evaporation of water =
`H_e=2.4 kJ/g`

Mass of water evaporated , m= 60 g

Amount of heat required to evaporate 60 grams of water :Q

`Q = H_e* m=2.4 kJ/mol* 60 =144 kJ`

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 =
`T_1=35^oC=308.15 K`

Final temperature =
`T_2=20^oC=293.15 K`

Change in temperature =
`\Delta T=T_2-T_2`

`Q'=mc\Delta T`

`m=(Q')/(c\Delta T)=(-144,000 J)/(4.18 J/g K* (293.15 K-308.15 K))`

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.