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 \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K} .)\)

Short answer

1. Calculate \(Q_\mathrm{evap} = 60 \mathrm{~g} \cdot 2.4 \mathrm{~kJ/g} = 144 \mathrm{~kJ}\) 2. Convert \(Q_\mathrm{evap}\) to joules: \(144 \mathrm{~kJ} \cdot 1000 = 144,000 \mathrm{~J}\) 3. Calculate \(\Delta T = 35^\circ \mathrm{C} - 20^\circ \mathrm{C} = 15^\circ \mathrm{C}\) 4. Use the mass formula to find \(m\): \(m = \frac{144,000 \mathrm{~J}}{4.18 \mathrm{~J/gK} \cdot 15^\circ \mathrm{C}} \approx 2290\,\mathrm{g}\) Thus, approximately \(2290 \mathrm{~g}\) 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-by-step solution

Calculate the heat absorbed by the evaporated water

To find the total heat absorbed by the evaporated water, we need to multiply the mass of the evaporated water (60 g) by its heat of vaporization (\(2.4 \mathrm{~kJ} / \mathrm{g}\)): $$ Q_\mathrm{evap} = m_\mathrm{evap} \cdot L_\mathrm{v} $$ where \(Q_\mathrm{evap}\) is the heat absorbed by the evaporated water, \(m_\mathrm{evap}=60\,\mathrm{g}\) is the mass of the evaporated water, and \(L_\mathrm{v}=2.4\,\mathrm{kJ/g}\) is the heat of vaporization of water.

Convert the heat absorbed into joules

Since the specific heat of water is given in joules, it is necessary to convert the heat absorbed by the evaporated water from kilojoules to joules: $$ Q_\mathrm{evap(J)} = Q_\mathrm{evap(kJ)} \cdot 1000 $$

Calculate the mass of cooled water

Now, we'll use the heat absorbed by the evaporated water to find out how many grams of water can be cooled from \(35^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\). To do this, we use the formula: $$ Q = m \cdot c \cdot \Delta T $$ where \(Q\) is the heat absorbed, \(m\) is the mass of cooled water, \(c=4.18 \,\mathrm{J/gK}\) is the specific heat of water, and \(\Delta T = 35^{\circ}\mathrm{C} - 20^{\circ}\mathrm{C}\) is the change in temperature. Rearrange the formula to find the mass of cooled water: $$ m = \frac{Q}{c \cdot \Delta T} $$

Combine the steps and solve for m

From steps 1-3, we have all the necessary information to solve for \(m\). Plug the values into the formula from step 3 and solve for the mass of cooled water: 1. Calculate \(Q_\mathrm{evap} = 60 \mathrm{~g} \cdot 2.4 \mathrm{~kJ/g}\) 2. Convert \(Q_\mathrm{evap}\) to joules 3. Calculate \(\Delta T = 35^\circ \mathrm{C} - 20^\circ \mathrm{C}\) 4. Use the mass formula to find \(m\) After solving for \(m\), we will have the mass of water that can be cooled from \(35^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\) by the evaporation of \(60 \mathrm{~g}\) of water.

Key concepts

Specific Heat Capacity

Specific heat capacity is a fundamental concept in understanding heat transfer. It represents the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). In our exercise, water has a specific heat capacity of \(4.18 \mathrm{~J/gK}\). This value is crucial because it tells us how much energy is needed to alter the temperature of water.
For example, if we wanted to know how much heat is needed to change the temperature of 1 gram of water by 15 degrees Celsius, we would multiply the mass (1 g) by the specific heat capacity (4.18 J/gK) and the change in temperature (15 K).
This concept is particularly important when dealing with thermal processes in a variety of scientific and engineering applications, including climate science, thermal engineering, and everyday phenomena like cooking and heating systems.
Understanding specific heat capacity allows us to calculate precisely how substances can absorb or release energy, enabling precise control over temperature changes and energy usage.

Heat of Vaporization

Heat of vaporization is the amount of energy required to transform a given quantity of a liquid into a gas at a constant temperature. This is a crucial property of substances and plays a significant role in heat transfer, especially when it comes to boiling or condensation processes.
In the context of our exercise, water requires \(2.4 \mathrm{~kJ/g}\) to vaporize. This indicates that each gram of water needs 2.4 kilojoules of energy to go from a liquid to a gaseous state without a change in temperature. The heat absorbed is utilized to overcome the intermolecular forces within the liquid, facilitating its transformation into vapor. This process involves a significant energy transfer which is often exploited in cooling mechanisms and steam generation.
Applications of the heat of vaporization include distillation, power generation through steam turbines, and even natural phenomena like the water cycle. Understanding this concept is essential to grasp how energy changes states of matter and contributes to various thermal processes across multiple contexts.

Evaporative Cooling

Evaporative cooling is a phenomenon where the evaporation of a liquid, typically water, leads to a cooling effect on the remaining liquid. This is due to the fact that when a liquid evaporates, it takes away with it the latent heat of vaporization, thereby lowering the temperature of the remaining liquid.
In our exercise, as 60 grams of water evaporate, they absorb a specific amount of heat energy corresponding to the heat of vaporization. This energy absorption results in a cooling effect on the water left behind, demonstrating how evaporative cooling occurs naturally.
Evaporative cooling is a ubiquitous process in nature and technology, used in applications ranging from cooling water in clay pots in arid regions to sophisticated modern air conditioning systems. It is an efficient way to reduce temperatures without extensive energy input, relying instead on the natural properties of water and heat transfer.
Understanding how evaporative cooling works provides insights into more sustainable and cost-effective ways to manage heat and thermal processes in various environments.