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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?

Short Answer

Expert verified
(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 by step solution

01

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 \( c = 4.186 \, \text{kJ/kg} \, ^{\circ}\text{C} \).
02

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:\[Q = 0.250 \, \text{kg} \times 4.186 \, \text{kJ/kg} \, ^{\circ}\text{C} \times (20.0^{\circ} \text{C} - 85.0^{\circ} \text{C}) = -68.37 \, \text{kJ}\]
03

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:\[\Delta S_{\text{water}} = \frac{-68.37}{325.65} = -0.210 \, \text{kJ/K}\]
04

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 \( T_{\text{air}} = 20.0^{\circ} \text{C} = 293.15 \, \text{K} \):\[\Delta S_{\text{air}} = \frac{+68.37}{293.15} = +0.233 \, \text{kJ/K}\]
05

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:\[\Delta S_{\text{total}} = \Delta S_{\text{water}} + \Delta S_{\text{air}}\]\[\Delta S_{\text{total}} = -0.210 + 0.233 = +0.023 \, \text{kJ/K}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

20.37. You design a Carnot engine that operates between temperatures of 500 \(\mathrm{K}\) and 400 \(\mathrm{K}\) and produces 2000 \(\mathrm{J}\) of work in each cycle. (a) Calculate your engine's efficiency. (b) Calculate the amount of heat discarded during the isothermal compression at 400 \(\mathrm{K}\) . (c) Sketch the 500 \(\mathrm{K}\) and 400 \(\mathrm{K}\) isotherms on apV-diagram (no calculations); then sketch the Carnot cycle followed by your engine. (d) On the same diagram, sketch the 300 \(\mathrm{K}\) isotherm; then sketch, in a different color if possible, the Carnot cycle starting at the same point on the 500 \(\mathrm{K}\) isotherm but operating in a cycle between the 500 \(\mathrm{K}\) and 300 \(\mathrm{K}\) isotherms. (e) Compare the areas inside the loops (the net work done) for the two cycles. Notice that the same amount of heat is extracted from the hot reservoir in both cases. Can you explain why less heat is "wasted" during the 300 \(\mathrm{K}\) isothermal compression than during the 400 \(\mathrm{K}\) compression?

A Gasoline Engine. Agasoline engine takes in \(1.61 \times 10^{4} \mathrm{J}\) of heat and delivers 3700 \(\mathrm{J}\) of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of \(4.60 \times 10^{4} \mathrm{J} / \mathrm{g}\) . (a) What is the thermal efficiency? (b) How much heat is discarded in each cycle? (c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower?

20.12. A freezer has a coefficient of performance of \(2.40 .\) The freezer is to convert 1.80 \(\mathrm{kg}\) of water at \(25.0^{\circ} \mathrm{C}\) to 1.80 \(\mathrm{kg}\) of ice at \(-5.0^{\circ} \mathrm{C}\) in hour. (a) What amount of heat must be removed from the water at \(25.0^{\circ} \mathrm{C}\) to convert it to ice at \(-5.0^{\circ} \mathrm{C} ?\) (b) How much electrical energy is consumed by the freezer during this hour? (c) How much wasted heat is delivered to the room in which the freezer sits?

20.34 . A box is separated by a partition into two parts of equal volume. The left side of the box contains 500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature. The partition is punctured, and equilibrium is eventually attained. Assume that the volume of the box is large enough for each gas to undergo a free expansion and not change temperature, (a) On average, how many molecules of each type will there be in either half of the box? (b) What is the change in entropy of the system when the partition is punctured? the (c) What is the probability that the molecules will be found in the same distribution as they were before the partition was punctured - that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right half?

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?

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