Problem 28 - University Physics with Modern Physics

Question

20.28. You make tea with 0.250 \(\mathrm{kg}\) of \(85.0^{\circ} \mathrm{C}\) water and let it cool to room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) before drinking it. (a) Calculate the entropy change of the water while it cools. (b) The cooling process is essentially isothernal for the air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all the heat lost by the water goes into the air. What is the total entropy change of the system tea \(+\) air?

Step-by-Step Solution

Answer

(a) \( \Delta S_{\text{water}} = -0.210 \, \text{kJ/K} \), (b) \( \Delta S_{\text{air}} = +0.233 \, \text{kJ/K} \). Total \( \Delta S_{\text{total}} = +0.023 \, \text{kJ/K} \).

Step 1: Identify Known Values

We are given that the mass of the water is \( m = 0.250 \, \text{kg} \), its initial temperature \( T_i = 85.0^{\circ} \text{C} \), and its final temperature \( T_f = 20.0^{\circ} \text{C} \). The specific heat capacity of water is \( c = 4.186 \, \text{kJ/kg} \, ^{\circ}\text{C} \).

Step 2: Calculate Heat Lost by Water

The heat lost by the water as it cools can be calculated using the formula \( Q = m \cdot c \cdot (T_f - T_i) \). Substitute the known values:\[Q = 0.250 \, \text{kg} \times 4.186 \, \text{kJ/kg} \, ^{\circ}\text{C} \times (20.0^{\circ} \text{C} - 85.0^{\circ} \text{C}) = -68.37 \, \text{kJ}\]

Step 3: Calculate Entropy Change of Water

The entropy change \( \Delta S \) for the water can be calculated using the formula:\[\Delta S = \int_{T_i}^{T_f} \frac{dQ}{T}\]For constant pressure, this simplifies to:\[\Delta S = \frac{Q}{T_{\text{avg}}}\]Where \( T_{\text{avg}} = \frac{T_i + T_f}{2} = \frac{85.0 + 20.0}{2} = 52.5 \, ^{\circ}\text{C} = 325.65 \, \text{K} \).Substitute into the formula:\[\Delta S_{\text{water}} = \frac{-68.37}{325.65} = -0.210 \, \text{kJ/K}\]

Step 4: Entropy Change of Air

Since all heat lost by the water is absorbed by the air, the change in entropy of the air is:\[\Delta S_{\text{air}} = \frac{+68.37}{T_{\text{air}}}\]Given that the air is at room temperature \( T_{\text{air}} = 20.0^{\circ} \text{C} = 293.15 \, \text{K} \):\[\Delta S_{\text{air}} = \frac{+68.37}{293.15} = +0.233 \, \text{kJ/K}\]

Step 5: Calculate Total Entropy Change

The total entropy change of the system \( \Delta S_{\text{total}} \) is the sum of the entropy changes of the water and the air:\[\Delta S_{\text{total}} = \Delta S_{\text{water}} + \Delta S_{\text{air}}\]\[\Delta S_{\text{total}} = -0.210 + 0.233 = +0.023 \, \text{kJ/K}\]

Key Concepts

Entropy Change

In thermodynamics, entropy is a measure of the amount of energy in a physical system that is not available to do work. Entropy changes when heat is transferred between systems or within a single system. When you have a system cooling or warming, such as your tea cooling in the kitchen, it involves a change in entropy.

The entropy change for a system, like the water in the tea, can be approached using the integral formula: \[\Delta S = \int \frac{dQ}{T}\]However, for processes where the system's temperature changes uniformly, like cooling water, this can be simplified using: \[\Delta S = \frac{Q}{T_{\text{avg}}}\]Here, \(Q\) is the heat transfer amount, and \(T_{\text{avg}}\) is the average temperature in Kelvin. This is useful when calculating the entropy change as tea cools down, transferring heat to the air.

In the example, we calculated the entropy change of water during its cooling phase, resulting in \(-0.210 \, \text{kJ/K}\). The negative sign reflects entropy going from the system into the surrounding air.

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics involving the movement of thermal energy from one object or material to another due to a temperature difference. It's essential for understanding how heat leaves the hot tea and is gained by the surrounding air. As the tea cools, heat is transferred from the tea (the hot system) to the air (the cooler system) until thermal equilibrium is reached.

The formula for calculating the amount of heat transferred when a substance changes temperature is: \[ Q = m \cdot c \cdot (T_f - T_i) \]where:

• \( Q \) is the heat transferred
• \( m \) is the mass of the substance
• \( c \) is the specific heat capacity
• \( T_f \) and \( T_i \) are the final and initial temperatures, respectively

In our scenario, by inserting the known values, we found that \(-68.37 \, \text{kJ}\) of heat left the water. The negative value indicates the direction of heat flow away from the water.

Specific Heat Capacity

Specific heat capacity is crucial for understanding how substances store heat. It represents the amount of heat per unit mass required to raise the temperature of a substance by one degree Kelvin or Celsius. Water, for instance, has a specific heat capacity of \(4.186 \, \text{kJ/kg} \, ^{\circ}\text{C}\), making it particularly efficient at storing thermal energy. This explains why it takes longer to heat or cool than many other substances.

In practical terms, specific heat capacity allows us to predict how much heat is needed to change a material's temperature, or conversely, how much temperature change will occur when a specific amount of heat is absorbed or lost. In the tea-cooling example:

• We calculated the total heat loss of water as it cooled
• Using specific heat capacity allows us to determine this heat exchange accurately

Understanding these principles lets us not only solve textbook problems but also comprehend energy efficiency and heat management in real-world applications.