Problem 24 - University Physics with Modern Physics

Question

You make tea with 0.250 kg of 85.0\(^\circ\)C water and let it cool to room temperature (20.0\(^\circ\)C). (a) Calculate the entropy change of the water while it cools. (b) The cooling process is essentially isothermal for the air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all of 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

The total entropy change of the system is approximately 142 J/K.

Step 1: Identify the Known Values

We have initial temperature \(T_i = 85.0\, ^\circ\text{C} = 358.15\, \text{K}\) and final temperature \(T_f = 20.0\, ^\circ\text{C} = 293.15\, \text{K}\). The mass of water is \(m = 0.250\, \text{kg}\). The specific heat capacity of water is \(c = 4186\, \text{J/(kg\cdot K)}\).

Step 2: Calculate the Change in Entropy of Water

The change in entropy \(\Delta S_w\) of the water as it cools can be modeled as an isochoric process using the formula:\[\Delta S_w = m \cdot c \cdot \ln \left( \frac{T_f}{T_i} \right)\]Substitute the known values:\[\Delta S_w = 0.250 \times 4186 \times \ln \left( \frac{293.15}{358.15} \right)\]Calculate \(\Delta S_w\).

Step 3: Solve for Water Entropy Change

Compute the numerical value:\[\Delta S_w = 0.250 \times 4186 \times \ln \left( \frac{293.15}{358.15} \right) \approx -89.6\, \text{J/K}\]

Step 4: Calculate the Heat Lost by Water

To find the heat lost by water, use the formula:\[Q = m \cdot c \cdot (T_f - T_i)\]Substitute the values:\[Q = 0.250 \times 4186 \times (293.15 - 358.15)\]Calculate \(Q\).

Step 5: Solve for Heat Lost by Water

Compute the numerical value:\[Q = 0.250 \times 4186 \times (293.15 - 358.15) \approx -67907.5\, \text{J}\]

Step 6: Calculate the Entropy Change of the Air

Since the air receives the heat \(Q\) and is assumed to keep constant temperature (isothermal process), the change in entropy \(\Delta S_a\) for the air is given by:\[\Delta S_a = \frac{Q}{T_a}\]where \(T_a = 293.15\, \text{K}\). Substitute \(Q\):\[\Delta S_a = \frac{67907.5}{293.15}\]Calculate \(\Delta S_a\).

Step 7: Solve for Air Entropy Change

Compute the numerical value:\[\Delta S_a = \frac{67907.5}{293.15} \approx 231.6\, \text{J/K}\]

Step 8: Calculate the Total Entropy Change of the System

The total entropy change for the system \(\Delta S_{total}\) is the sum of the entropy changes of the water and the air:\[\Delta S_{total} = \Delta S_w + \Delta S_a\]Substitute the calculated values:\[\Delta S_{total} = -89.6 + 231.6 \approx 142\, \text{J/K}\]

Step 9: Final Result: Total Entropy Change

The total entropy change of the system (tea + air) is approximately:\[142\, \text{J/K}\]

Key Concepts

Thermodynamics

In the realm of physics, thermodynamics is a branch that focuses on the study of heat, work, and energy. One of its central concepts is entropy, a measure of disorder or randomness in a system. Understanding how entropy changes during various processes helps us understand how energy transformations impact our world.
Thermodynamics is governed by several laws:

• The First Law of Thermodynamics states that energy cannot be created or destroyed, only transformed.”
• The Second Law of Thermodynamics introduces the concept of entropy, stating that in an isolated system, the total entropy can never decrease over time.

As we apply these principles, we get to explore how energy transition affects systems such as engines, refrigerators, and in our case, a cooling cup of tea. Understanding these principles provides insight into energy efficiency, the feasibility of processes, and even the limitations of energy usage.

Heat Transfer

Heat transfer is the movement of thermal energy from one object or material to another. It occurs through three main mechanisms: conduction, convection, and radiation. In our exercise, it involves the heat lost by water as it cools and the corresponding heat gain by the air around it.

• **Conduction:** This is the transfer of heat through a solid material, like the spoon in our tea.
• **Convection:** This is the transfer through a fluid (air or liquid) and is typically responsible for the cooling effect when the tea gives up heat to the surrounding air.
• **Radiation:** This is energy transfer through electromagnetic waves and can also be a minor factor in the cooling of tea but often overshadowed by convection in this scenario.

The formula for heat transfer as seen in the exercise, involves the specific heat capacity of water, which is the amount of energy required to change the temperature of one kilogram of water by one Kelvin. This understanding can be extended to calculate energy requirements for heating or cooling other substances and systems.

Isothermal Process

An isothermal process is one that occurs at a constant temperature. In this scenario, as the tea cools, the kitchen air is treated as being constant in temperature, making it an isothermal system for calculation purposes.
In the context of this exercise, because the air temperature doesn't change, the entropy change for the air can be calculated directly by dividing the heat absorbed by the constant temperature of the air. This simplification is crucial in solving practical problems as it avoids intricate temperature variable integrations.
The formula used here, \[\Delta S_a = \frac{Q}{T_a}\] illustrates this point as it directly relates the heat exchange to entropy change through constant temperature. Such processes are vital in understanding thermal cycles, and implications in broad applications like heat engines and refrigerators, where maintaining constant temperature enhances efficiency.