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Chapter 11: Problem 43

For many years drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from 35 to \(20^{\circ} \mathrm{C}\) by the evaporation of \(60 \mathrm{~g}\) of water? (The heat of vaporization of water in this temperature range is \(2.4 \mathrm{~kJ} / \mathrm{g} .\) The specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K} .)\)

Short Answer

Expert verified
The mass of water that can be cooled from 35°C to 20°C by evaporating 60g of water is approximately 2294 grams.

Step by step solution

01

Calculate the heat absorbed during evaporation

To calculate the heat absorbed during evaporation, we can use the equation: Q_evaporation = Mass of evaporated water × Heat of vaporization where Mass of evaporated water = 60g Heat of vaporization = 2.4 kJ/g First, convert the heat of vaporization from kJ/g to J/g: \(2.4 \frac{kJ}{g} = 2400 \frac{J}{g}\) Now, we can calculate Q_evaporation: \(Q_{evaporation} = 60 g \times 2400 \frac{J}{g} = 144000 J\)
02

Calculate the heat needed to cool the water

To calculate the heat needed to cool the water, we can use the equation: Q_cooling = Specific heat of water × Mass of water to be cooled × Temperature difference where Specific heat of water = 4.18 J/g-K Mass of water to be cooled = m (unknown) Temperature difference = 35°C - 20°C = 15 K Now, we can write the equation for Q_cooling: \(Q_{cooling} = 4.18 \frac{J}{g-K} \times m \times 15 K\)
03

Equate Q_evaporation and Q_cooling, and solve for m

Since the heat absorbed during evaporation must equal the heat needed to cool the water, we can write: \(144000 J = 4.18 \frac{J}{g-K} \times m \times 15 K\) Now, solve for m: \(m = \frac{144000 J}{4.18 \frac{J}{g-K} \times 15 K} = \frac{144000}{62.7} g \approx 2294 g\) The mass of water that can be cooled from 35°C to 20°C by evaporating 60g of water is approximately 2294 grams.

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Most popular questions from this chapter

The vapor pressure of a volatile liquid can be determined by slowly bubbling a known volume of gas through it at a known temperature and pressure. In an experiment, \(8.00 \mathrm{~L}\) of argon gas is passed through $11.7872 \mathrm{~g}\( of liquid hexane \)\mathrm{C}_{6} \mathrm{H}_{14}$ at \(30.0^{\circ} \mathrm{C}\). The mass of the remaining liquid after the experiment is \(4.875 \mathrm{~g}\). Assuming that the gas becomes saturated with hexane vapor and that the total gas volume and temperature remain constant, what is the vapor pressure of hexane in atm?

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