Problem 37 - Chemistry: The Central Science

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 \(35^{\circ} \mathrm{C}\) 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 \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K} .)\)

Step-by-Step Solution

Answer

\(2296 \mathrm{~g}\) of water can be cooled from \(35^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\) by the evaporation of \(60 \mathrm{~g}\) of water.

Step 1: We are given that \(60 \mathrm{~g}\) of water evaporates, and the heat of vaporization is \(2.4 \mathrm{~kJ} / \mathrm{g}\). First, we need to convert the heat of vaporization to Joules: Heat of vaporization = \(2.4 \mathrm{~kJ} / \mathrm{g} \times (1000 \mathrm{~J} / \mathrm{kJ}) = 2400 \mathrm{~J} / \mathrm{g}\) Now, we can calculate the amount of heat lost due to evaporation: Heat lost due to evaporation (Q) = (mass of evaporated water) \(\times\) (heat of vaporization) Q = \(60 \mathrm{~g} \times 2400 \mathrm{~J} / \mathrm{g} = 144000 \mathrm{~J}\) #Step 2: Calculate the amount of heat required to cool the remaining water#

To find out the amount of heat required to cool the remaining water from \(35^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\), we can use the specific heat formula: Heat required (Q) = (mass of water to be cooled) \(\times\) (specific heat of water) \(\times\) (change in temperature) Let's denote the mass of water to be cooled as 'm'. Then, the change in temperature is \(35^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 15^{\circ} \mathrm{C}\). The specific heat of water is given as \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K}\). Now, the formula becomes: Q = m \(\times 4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K} \times 15^{\circ} \mathrm{C}\) #Step 3: Equate the heat lost due to evaporation to the heat required to cool the remaining water#

Step 2: As the heat lost due to evaporation is used to cool the remaining water, we can equate the heat values we calculated in Step 1 and Step 2: \(144000 \mathrm{~J} = m \times 4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K} \times 15^{\circ} \mathrm{C}\) #Step 4: Calculate the mass of water that can be cooled#

Now, we can solve the equation in Step 3 for the mass (m) of water that can be cooled: \[m = \frac{144000 \mathrm{~J}}{4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K} \times 15^{\circ} \mathrm{C}}\] \[m = \frac{144000 \mathrm{~J}}{62.7 \mathrm{~J}/\mathrm{g}}\] \[m = 2296 \mathrm{~g}\] Hence, \(2296 \mathrm{~g}\) of water can be cooled from \(35^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\) by the evaporation of \(60 \mathrm{~g}\) of water.