Problem 29 - University Physics with Modern Physics

Question

20.29. 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) -234.2 J/K; (b) Air: 232.3 J/K, Total: -1.9 J/K.

Step 1: Understand the problem

We need to calculate the entropy change for both the water (tea cooling down) and the air in the room and combine these to find the total entropy change.

Step 2: Calculate entropy change of the water

The entropy change of the water can be calculated using the formula: \[ \Delta S = m \cdot c \cdot \ln\left(\frac{T_2}{T_1}\right) \]where \(m\) is the mass of the water (0.250 kg), \(c\) is the specific heat capacity of water (4.186 J/g°C), and \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin.Convert temperatures from Celsius to Kelvin by adding 273.15:\[ T_1 = 85.0 + 273.15 = 358.15 \, \text{K} \]\[ T_2 = 20.0 + 273.15 = 293.15 \, \text{K} \]Calculate \(\Delta S\):\[ \Delta S = 0.250 \cdot 4.186 \cdot 1000 \cdot \ln\left(\frac{293.15}{358.15}\right) \]\[ \Delta S \approx -234.2 \, \text{J/K} \]

Step 3: Calculate entropy change of the air

Assuming that all the heat lost by the water is gained by the air, the change in entropy of the air \(\Delta S_{air}\) can be calculated using:\[ \Delta S_{air} = \frac{Q_{lost}}{T_{air}} \]Calculate the heat lost by the water \(Q_{lost}\):\[ Q_{lost} = m \cdot c \cdot (T_2 - T_1) \]\[ Q_{lost} = 0.250 \cdot 4.186 \cdot 1000 \cdot (20.0 - 85.0) \]\[ Q_{lost} = -68107.5 \, \text{J} \] (use only magnitude for entropy change)The air temperature is assumed to be constant at room temperature \(T_{air} = 293.15 \, \text{K}\):\[ \Delta S_{air} = \frac{68107.5}{293.15} \]\[ \Delta S_{air} \approx 232.3 \, \text{J/K} \]

Step 4: Calculate total entropy change of the system

The total change in entropy of the system (tea + air) is the sum of the changes in entropy of the water and air:\[ \Delta S_{total} = \Delta S + \Delta S_{air} \]\[ \Delta S_{total} = -234.2 + 232.3 \]\[ \Delta S_{total} \approx -1.9 \, \text{J/K} \]

Step 5: Final Step: Interpret the result

The slightly negative total entropy change suggests a small decrease in the entropy of the system, which can happen locally due to the entropy exchange with the surroundings. However, in reality, the overall entropy of the universe increases.

Key Concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat, work, temperature, and energy. It's a vast field, but one of the foundational concepts is the tendency for processes to evolve towards a state of maximum entropy. Entropy is essentially a measure of disorder or randomness in a system. Thermodynamics has laws, much like traffic rules, that help us understand how energy transformations occur.

• The First Law of Thermodynamics states that energy cannot be created or destroyed, only transformed or transferred. This law is crucial when analyzing heat transfers, as any heat loss by one body must be gained by another.
• The Second Law of Thermodynamics introduces the concept of entropy, stating that in any energy transfer, some energy will dissipate as heat, increasing the overall entropy.
• The Third Law of Thermodynamics states that as the temperature of a perfect crystal approaches absolute zero, its entropy approaches a minimum value.

This framework allows us to quantify changes in energy, like in the given exercise, where we calculate the entropy change when tea cools down.

Heat Transfer

Heat transfer occurs when thermal energy is exchanged between physical systems. It can happen through various methods such as conduction, convection, or radiation.

• Conduction involves heat flowing through a material, typical of solids. It's like how heat from a hot stove will travel up a metal pot handle if it's left unattended.
• Convection refers to the transfer of heat by the movement of fluids or gases, such as when boiling water circulates to distribute heat evenly.
• Radiation is the transfer of heat through electromagnetic waves, like the warmth felt from sunlight.

In the tea cooling scenario, the heat transfer is predominantly due to conduction and convection. The water in the mug loses heat to the surrounding air, and this process continues until a thermal equilibrium is reached.

Specific Heat Capacity

Specific heat capacity is a property that describes how much heat is required to change the temperature of a given mass by one degree. It is unique to each material, indicating how they respond to heat.
For water, its specific heat capacity is relatively high (4.186 J/g°C). This means it requires more energy to change the temperature of water by one degree compared to many other substances. This property makes water an excellent medium for heat storage and transfer.

• When calculating changes in the water's temperature, like in the exercise, we use this capacity to determine how much heat is lost or gained, using the formula \[ Q = m \cdot c \cdot \Delta T \].
• This helps in calculating how much the temperature will drop (or rise), and thus assists in finding the entropy change associated with the process.

Isothermal Process

In thermodynamics, an isothermal process is one where the temperature remains constant throughout. This is especially significant in scenarios where heat transfer occurs but without changing the system's temperature.
In the context of the exercise, while the tea cools, it's assumed that the surrounding air behaves isothermally. Hence, any heat lost by the tea is gained by the air without changing its temperature.

• This assumption simplifies calculations as it allows us to take a constant temperature for the air when calculating its entropy change, using the formula \[ \Delta S = \frac{Q}{T} \].
• It reflects situations where thermal reservoirs are large enough to absorb small amounts of heat without significantly altering their own temperature.

Such processes showcase the elegance of thermodynamic analysis by stripping down complexities to manageable assumptions, providing clear insights into energy distributions.