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Ch.11 - Liquids and Intermolecular Forces
All textbooksBrown 15th EditionCh.11 - Liquids and Intermolecular ForcesProblem 45
Chapter 11, Problem 45

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 °C by the evaporation of 60 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).

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First, calculate the amount of heat absorbed by the evaporation of 60 g of water using the heat of vaporization. The formula is Q = m * Hv, where Q is the heat absorbed, m is the mass of the water, and Hv is the heat of vaporization.
Next, calculate the amount of heat released by the cooling of water from 35 to 20 °C using the specific heat formula. The formula is Q = m * c * ΔT, where Q is the heat released, m is the mass of the water, c is the specific heat, and ΔT is the change in temperature.
Since the heat absorbed by evaporation is equal to the heat released by cooling, you can set the two equations equal to each other and solve for the mass of the water that is cooled.
Substitute the given values into the equation and solve for the mass of the water.
Remember to check your units. The heat of vaporization is given in kJ/g, but the specific heat is given in J/g-K. You'll need to convert one to the other before you can solve the equation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert a unit mass of a liquid into vapor without a change in temperature. For water, this value is significant because it indicates how much heat energy is absorbed during the evaporation process. In this question, the heat of vaporization is given as 2.4 kJ/g, which will be crucial for calculating how much water can be cooled by the evaporation of 60 g of water.
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Heat Capacity

Specific Heat Capacity

Specific heat capacity is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). For water, the specific heat capacity is 4.18 J/g-K, meaning it takes 4.18 joules to raise the temperature of 1 gram of water by 1 °C. This concept is essential for determining how much heat energy is needed to cool the water from 35 °C to 20 °C in the problem.
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Heat Capacity

Energy Conservation in Phase Changes

Energy conservation in phase changes refers to the principle that energy is neither created nor destroyed but can change forms. In this context, the energy lost by the water being cooled must equal the energy gained by the evaporating water. This relationship allows us to set up an equation to find out how much water can be cooled by the evaporation of a specific mass of water, linking the heat of vaporization and the specific heat capacity.
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Entropy in Phase Changes
Related Practice
Textbook Question

Ethyl chloride (C2H5Cl) boils at 12 °C. When liquid C2H5Cl under pressure is sprayed on a room-temperature (25 °C) surface in air, the surface is cooled considerably. (a) What does this observation tell us about the specific heat of C2H5Cl(g) as compared with that of C2H5Cl(l)?

Textbook Question

Ethyl chloride (C2H5Cl) boils at 12 °C. When liquid C2H5Cl under pressure is sprayed on a room-temperature (25 °C) surface in air, the surface is cooled considerably. (b) Assume that the heat lost by the surface is gained by ethyl chloride. What enthalpies must you consider if you were to calculate the final temperature of the surface?