Question

For many years drinking water has been cooled in hot climates by evaporating it from the surface of canvas bags or porous clay pots.How many grams of water can be cooled from 40 ∘C to 20 ∘C by the evaporation of 43 g of water? (The heat of vaporization of water in this temperature range is 2.4 kJ/g. The specific heat of water is 4.18 J/g⋅K.)

Answer 1

Answer:

1595g

Explanation:

heat of vaporization=2.4 kk/g

mass of water to be vaporized=50g

heat realised=mass of water to be vaporized X heat of vaporization=

50g×2.4kj/g

= 120k

= 120000

initial temperature=33+273=306k

final temperature=15+273= 288k

change in temperature requi=T= 306-18k

specific heat of water is 4.18j/gk

mass of water that can be cooled=total heat/(specific heat of water x change in temperature)

120000j/(4.18j/gk×18k)

1595g