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

Approximately 2297.29 grams of water can be cooled from 35°C to 20°C by the evaporation of 60g of water.

Step-by-step solution

Compute the heat released upon evaporation of 60g of water

We have the heat of vaporization of water (2.4 kJ/g) and the mass of evaporated water (60 g). We can now calculate the heat released upon evaporation using the formula: Heat released (Q) = mass × heat of vaporization First, convert the heat of vaporization from kJ/g to J/g to be consistent with the units of the specific heat. (2.4 kJ/g) × (1000 J/1 kJ) = 2400 J/g Now calculate the heat released: Q = (60 g) × (2400 J/g) = 144,000 J

Calculate the temperature change

Compute the temperature change, ΔT, which is the difference between the initial and final temperatures: ΔT = Tf - Ti = 20°C - 35°C = -15°C

Compute the mass of water that can be cooled from 35°C to 20°C

To find how many grams of water can be cooled under these conditions, we need to determine the mass (m) of water, applying the specific heat equation: Q = m × specific heat × ΔT Rearrange the equation to solve for m: m = Q / (specific heat × ΔT) Since the temperature change is negative (cold), the heat flow is negative as well. We need to use the absolute values of ΔT and Q, which makes more sense physically: m = 144,000 J / (4.18 J/g-K × |-15°C|) m = 144,000 J / (4.18 J/g-K × 15 K) m = 144,000 J / 62.7 J/g m ≈ 2297.29 g So, approximately 2297.29 grams of water can be cooled from 35°C to 20°C by the evaporation of 60g of water.

Key concepts

Heat of Vaporization

The heat of vaporization is a term used to describe the amount of energy required to convert a unit mass of a substance from the liquid phase to the vapor phase at constant temperature and pressure. When a liquid evaporates, it absorbs heat from its surroundings, which leads to cooling. This principle is utilized in various cooling techniques, such as when water is evaporated from the surfaces of canvas bags to cool the drinking water within them.

In the context of the original exercise, the heat of vaporization is given as 2.4 kJ/g for water. This value is high, indicating the strong molecular forces between water molecules that must be overcome during evaporation. To find the total heat absorbed during the evaporation of water, you multiply the heat of vaporization by the mass of the water that evaporates. The higher the heat of vaporization, the more efficient the substance is in absorbing heat from its surroundings.

Understanding heat of vaporization is crucial because it not only dictates the amount of cooling that can be achieved but also provides insight into the intrinsic properties of a substance's phase change behavior.

Specific Heat

The specific heat of a substance is defined as the amount of heat per unit mass required to raise the temperature of that substance by one degree Celsius (or one Kelvin). The specific heat represents the heat capacity of a material and can vary depending on the state of the substance (solid, liquid, or gas) and its composition.

For water, the specific heat is approximately 4.18 J/g-K, which means it takes 4.18 joules of energy to raise the temperature of one gram of water by one degree Kelvin. This relatively high specific heat capacity of water makes it an excellent candidate for thermal regulation applications, from cooling systems to climate control.

In our exercise, specific heat is essential for calculating the mass of water that can be cooled by the evaporation process. It is a crucial factor in determining how much energy is required to change the temperature of a given amount of water.

Temperature Change

The temperature change in a substance, often denoted as \( \Delta T \), is a fundamental concept in thermodynamics. It represents the difference between the final temperature and the initial temperature of a system. In a cooling process, the temperature change is typically negative, indicating a drop in temperature.

In our example, the water in the canvas bags is initially at 35°C, and we aim to cool it to 20°C, resulting in a temperature change of \( \Delta T = 20°C - 35°C = -15°C \). When calculating the thermal energy transfer associated with this temperature change, we use the specific heat and mass of the substance, along with \( \Delta T \) to quantify the energy involved in this temperature reduction.

It is important to note that when using \( \Delta T \) in calculations involving heat exchange, one should always use its absolute value to ensure the physical significance of the result. Otherwise, the calculation could incorrectly indicate a gain in heat rather than a loss.

Enthalpy of Vaporization

The enthalpy of vaporization, also known as the latent heat of vaporization, is the heat required to transform a given amount of a substance from a liquid to a gas at constant pressure and temperature. It is a type of latent heat, meaning it's the energy used for the phase change without changing the temperature. This concept is vital in understanding processes like boiling, evaporation, and condensation.

The enthalpy of vaporization is a crucial factor when calculating the cooling effect of water evaporation, as seen in the exercise. It reflects the inherent energy change during the phase transition. Moreover, the connection between enthalpy and the heat of vaporization lies in the fact that the enthalpy of vaporization is essentially the heat of vaporization expressed per mole of substance instead of per gram.

Knowing the enthalpy of vaporization allows us to predict how much energy will be taken from the environment during evaporation, translating into how effectively a substance can be used for cooling purposes in applications such as the evaporative cooling of water in hot climates.