How many grams of water can be cooled from 33 ∘c to 15 ∘c by the evaporation of 50 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: -

1595 g

Explanation: -

Heat of vaporization = 2.4 Kj / g

Mass of water to be vaporized = 50 g

Heat released = Mass of water to be vaporized x Heat of vaporization

= 50 g x 2.4 KJ / g

= 120 KJ

= 120000 J

Initial temperature = 33+273 = 306 K

Final temperature = 15+273=288 K

Change in temperature required = T = 306 - 288 = 18 K

specific heat of water is 4.18 J / g K

Mass of water that can be cooled = Total heat / (specific heat of water x Change in temperature)

= 120000 J / (4.18 J / g K x 18 K)

= 1595 g