Problem 43 - Chemistry The Central Science In Si Units

Question

For many years drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags of 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 \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K}\).)

Step-by-Step Solution

Answer

Approximately 2296.49 grams of water can be cooled.

Step 1: Understand the Problem

We need to find out how many grams of water can be cooled from 35°C to 20°C using the evaporation of 60g of water. The heat required for this cooling will be provided by the vaporization of water.

Step 2: Calculate Heat Requirement for Cooling

The heat needed to cool a substance is calculated using the formula: \( q = mc\Delta T \), where \( m \) is the mass of the water being cooled, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Here, \( c = 4.18 \, \text{J/g°C} \) and \( \Delta T = 15 \, \text{°C} \). So, the heat required for cooling \( m \) grams of water is \( q = m \times 4.18 \times 15 \).

Step 3: Determine Heat from Evaporation

The heat released by the evaporation of 60g of water can be calculated using the formula: \( q = mL \), where \( L \) is the heat of vaporization. Here, \( m = 60 \, \text{g} \) and \( L = 2.4 \, \text{kJ/g} \). Convert \( L \) to \( \text{J/g} \): \( L = 2400 \, \text{J/g} \). Thus, the total heat from evaporation is \( 60 \times 2400 \).

Step 4: Set the Equations Equal

Set the heat gained by the cooling water equal to the heat given off by the evaporating water: \( m \times 4.18 \times 15 = 60 \times 2400 \).

Step 5: Solve for Mass of Water Cooled

Solve the equation \( m \times 4.18 \times 15 = 60 \times 2400 \) for \( m \). Simplifying gives: \( m = \frac{60 \times 2400}{4.18 \times 15} \). Calculate \( m \).

Step 6: Calculate the Mass of Water Cooled

Perform the calculation: \( m = \frac{144000}{62.7} \approx 2296.49 \). Thus, approximately 2296.49 grams of water can be cooled.

Key Concepts

Specific Heat Capacity

Specific heat capacity is a key concept when dealing with temperature changes in substances. It is the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius.
In our exercise, the specific heat capacity of water is given as 4.18 J/g°C. This means that to raise the temperature of 1 gram of water by 1°C, 4.18 joules of energy are needed.
The formula used to determine the heat needed for changing the temperature of a substance is:

• \( q = mc\Delta T \)

Here,

• \( q \) is the heat absorbed or released,
• \( m \) is the mass,
• \( c \) is the specific heat capacity,
• \( \Delta T \) is the temperature change.

By using this formula, we can calculate how much energy is required to cool down or heat up a specific amount of water.

Heat of Vaporization

The heat of vaporization is the amount of heat needed to turn a liquid into a gas without a change in temperature. It plays a crucial role in processes involving phase changes.
In the given exercise, the heat of vaporization for water is specified as 2.4 kJ/g.
When water evaporates, it absorbs a significant amount of energy from its surroundings, leaving the environment cooler. This is how evaporative cooling devices, like the clay pots mentioned in the exercise, work.

• The formula to calculate the energy involved in vaporization is:

• \( q = mL \)

where:

• \( q \) is the heat absorbed or released,
• \( m \) is the mass of the substance undergoing the phase change,
• \( L \) is the heat of vaporization.

In our example, by evaporating 60 grams of water, we use its heat of vaporization to extract heat from the surrounding water, causing it to cool down.

Temperature Change in Substances

Temperature change in substances relies heavily on specific heat capacity and heat transfer processes. When a substance gains or loses energy, it experiences a change in temperature unless it's undergoing a phase change.
For the cooling process described in the exercise, the temperature of some water is reduced from 35°C to 20°C, which is a change (\( \Delta T \)) of 15°C.
This temperature change occurs because the heat absorbed during the vaporization of another batch of water is transferred away from the liquid being cooled, effectively reducing its thermal energy.
Using the equation \( q = mc\Delta T \), we establish the connection between the energy transferred and the resulting temperature change. It becomes easy to see how much heat is absorbed from the surroundings to cool the water.
This process is the key to understanding why certain pressures and energies are needed to control the temperatures of substances in everyday life.

Thermodynamics in Chemistry

Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy, and by extension, of the relationship between all forms of energy.
In chemistry, understanding thermodynamics allows us to explain how reactions and physical changes occur, such as the evaporation and cooling of water in our exercise.
The Law of Conservation of Energy is fundamental, stating that energy cannot be created or destroyed but only converted from one form to another. This principle is showcased when heat from the water evaporation process is used to cool down another portion of water.
It means the energy involved in vaporizing one set of water molecules effectively reduces the energy or heat in the cooled mass, lowering its temperature.

• Key implications include understanding:
• Energy transformations during phase changes,
• Efficient energy management in systems,
• The practical use of energy conservation in chemical experiments and industrial processes.

All these relate back to controlling and implementing thermodynamic principles in systematic and beneficial ways.