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).

Key Concepts

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.

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.

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.