Question

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

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

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: \(Q_{evaporation} = 60 g \times 2400 \frac{J}{g} = 144000 J\)

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: \(Q_{cooling} = 4.18 \frac{J}{g-K} \times m \times 15 K\)

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.

Key concepts

Heat of Vaporization

When dealing with the heat of vaporization, we're talking about the energy required to convert a liquid into a gas at a constant temperature and pressure. Water, for instance, requires a significant amount of heat to change from its liquid state to a gaseous state. This is because the molecules need to break free from the forces holding them together in the liquid. In our problem, the heat of vaporization for water is given as 2.4 kJ/g. This is converted to 2400 J/g for ease of calculation.

Knowing the heat of vaporization is crucial when calculating how much energy is absorbed by evaporation. In our exercise, 60 g of water evaporates, absorbing 144,000 Joules of energy. This energy absorption is what enables water cooling through evaporation.

Specific Heat

Specific heat is the measure of heat energy required to raise the temperature of a given mass of a substance by one degree Celsius. Water has a specific heat value of 4.18 J/g-K, meaning it requires 4.18 Joules to increase the temperature of one gram of water by one degree Kelvin.

This characteristic shows why water is such an effective medium for temperature regulation: it takes more energy to increase its temperature compared to many other substances. In our exercise, the specific heat helps us calculate the energy needed to cool down water from 35°C to 20°C. This cooling energy must match the energy absorbed by the evaporating water for the system to balance.

Evaporation Cooling

Evaporation cooling is an exciting concept that leverages natural processes for temperature regulation. When water evaporates, it absorbs heat from its surroundings. This heat absorption leads to cooling, as seen in porous pots or canvas bags used in hot climates.

This method works because of the high heat of vaporization of water. In the exercise, we see an application where 60 g of water evaporates, absorbing heat and allowing 2294 g of water to cool from 35°C to 20°C. Such natural air conditioning showcases the power of thermodynamics and how simply altering the states of a substance can impact temperatures.