/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 You make tea with 0.250 \(\mathr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 isothermal for the air in your kitchen. Calculate the change in entropy of the air while the tea cols, 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) \( \Delta S_{water} \approx -90.23 \text{ J/K}\); (b) \( \Delta S_{air} \approx 232.1 \text{ J/K}\); Total: \( \Delta S_{total} \approx 141.87 \text{ J/K}\).

Step by step solution

01

Calculate Entropy Change of Water

The formula to calculate the change in entropy \( \Delta S \) for a substance changing temperature is given by \[ \Delta S = m c \ln \left( \frac{T_f}{T_i} \right) \] where \( m \) is the mass of the water (0.250 kg), \( c \) is the specific heat capacity of water (4.186 J/g°C), \( T_f \) is the final temperature (20.0°C), and \( T_i \) is the initial temperature (85.0°C). First convert temperatures to Kelvin: \( T_i = 358.15 \ \text{K} \), \( T_f = 293.15 \ \text{K} \). Then, calculate: \[ \Delta S = 0.250 \times 1000 \times 4.186 \times \ln \left( \frac{293.15}{358.15} \right) \] \[ \Delta S \approx 0.250 \times 4186 \times \ln(0.818) \approx -90.23 \, \text{J/K} \].
02

Calculate Entropy Change of Air

Since the process for the air is assumed to be isothermal, the change in entropy for the air is given by \[ \Delta S = \frac{Q}{T} \] where \( Q \) is the heat absorbed by the air (which is equal to the heat lost by the water when it cools) and \( T \) is the constant temperature of the air (approximately in Kelvin). The heat lost \( Q \) is given by \[ Q = m c (T_i - T_f) \] Substituting, we find \[ Q = 0.250 \times 1000 \times 4.186 \times (85.0 - 20.0) \approx 68027.5 \, \text{J} \] Assuming the air remains at room temperature, \( T = 293.15 \, \text{K} \), then \[ \Delta S = \frac{68027.5}{293.15} \approx 232.1 \, \text{J/K} \].
03

Calculate Total Entropy Change

The total entropy change of the system \( \Delta S_{total} \) is the sum of the entropy change of the water and the air: \[ \Delta S_{total} = \Delta S_{water} + \Delta S_{air} = -90.23 + 232.1 = 141.87 \, \text{J/K} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It describes how energy moves and changes in a system. The two main laws that often come into play are the first and the second laws of thermodynamics.
- The **first law** clarifies that energy cannot be created or destroyed; it can only change from one form to another.
- The **second law** introduces entropy, indicating that in a closed system, processes evolve in a direction that increases the total entropy.

In our tea-cooling scenario, these principles tell us how heat moves from the hot water to the air until both reach the same temperature. This heat transfer results in changes in the system's entropy, which we can calculate and understand through these thermodynamic laws.
Specific Heat Capacity
Specific heat capacity is an important concept that tells us how much energy is needed to raise the temperature of a unit mass of a substance by one degree Celsius. For water, this value is relatively high, 4.186 J/g°C, meaning water requires a lot of energy to change its temperature.
This property of water is why it can absorb a lot of heat without a big temperature change, which in turn reflects in its entropy calculations when it cools down. In practical terms, this means when you pour hot water to make tea, it can hold onto that heat longer, slowly releasing it into the surroundings such as the room air, thereby affecting the overall entropy.
Isothermal Processes
An isothermal process is one in which the temperature remains constant. In our scenario, although the temperature of the tea changes, the air's temperature remains stable, making it an isothermal process for the surrounding air.
For isothermal processes, the change in entropy (\( \Delta S \)) can be calculated by dividing the heat transfer (\( Q \)) by the constant temperature (\( T \)). This equation simplifies the calculation as we assume the environment's temperature does not fluctuate.

Even though the tea cools down, losing energy, the constant temperature of the air allows us to calculate how the surrounding air absorbs this heat, changing its entropy as a whole.
Heat Transfer
Heat transfer is the movement of thermal energy from a region of higher temperature to one of lower temperature, which in our case, occurs from the hot tea to the cooler kitchen air.
This process continues until thermal equilibrium is reached, where both the tea and the room air share a common temperature. Understanding heat transfer is essential to comprehend how and why the entropy of the system changes.
In our exercise, the heat lost from the water (tea) is absorbed by the surrounding air, confirming our calculation that looks at how entropy changes both for the tea as it cools and for the air absorbing the heat.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Heat Pump. A heat pump is a heat engine run in reverse. In winter it pumps heat from the cold air outside into the warmer air inside the building, maintaining the building at a comfortable temperature. In summer it pumps heat from the cooler air inside the building to the warmer air outside, acting as an air conditioner. (a) If the outside temperature in winter is \(-5.0^{\circ} \mathrm{C}\) and the inside temperature is \(17.0^{\circ} \mathrm{C}\) , how many joules of heat will the heat pump deliver to the inside for each joule of electrical energy used to run the unit, assuming an ideal Carnot cycle? (b) Suppose you have the option of using electrical resistance heating rather than a heat pump. How much electrical energy would you need in order to deliver the same amount of heat to the inside of the house as in part (a)? Consider a Carnot heat pump delivering heat to the inside of a house to maintain it at \(68^{\circ} \mathrm{F}\) . Show that the heat pump delivers less heat for each joule of electrical energy used to operate the unit as the outside temperature decreases. Notice that this behavior is opposite to the dependence of the efficiency of a Carnot heat engine on the difference in the reservoir temperatures. Explain why this is so.

You are designing a Carnot engine that has 2 \(\mathrm{mol}\) of \(\mathrm{CO}_{2}\) as its working substance; the gas may be treated as ideal. The gas is to have a maximum temperature of \(527^{\circ} \mathrm{C}\) and a maximum pressure of 5.00 atm. With a heat input of 400 \(\mathrm{J}\) per cycle, you want 300 \(\mathrm{J}\) of useful work. (a) Find the temperature of the cold reservoir. (b) For how many cycles must this engine run to melt completely a 10.0 -kg block of ice originally at \(0.0^{\circ} \mathrm{C},\) using only the heat rejected by the engine?

A Stirling-Cycle Engine. The Stirling cycle is similar to the Otto cycle, except that the compression and expansion of the gas are done at constant temperature, not adiabatically in the Otto cycle. The Stirling cycle is used in external combustion engines (in fact, burning fuel is not necessary; any way of producing a temperature difference will do-solar, geothermal, ocean temperature gradient, etc.), which means that the gas inside the cylinder is not used in the combustion process. Heat is supplied by burning fuel steadily outside the cylinder, instead of explosively inside the cylinder as in the Otto cycle. For this reason Stirling-cycle engines are quieter than Otto-cycle engines, since there are no intake and exhaust valves (a major source of engine noise). While small Stirling engines are used for a variety of purposes, Stirling engines for automobiles have not been successful because they are larger, heavier, and more expensive than conventional automobile engines. In the cycle, the working fluid goes through the following sequence of steps (Fig. P20.52): \begin{equation}\begin{array}{l}{\text { (i) Compressed isothermally at temperature } T_{1} \text { from the ir }} \\ {\text { state } a \text { to state } b, \text { with a compression ratio } r .} \\ {\text { (ii) Heated at constant volume to state } c \text { at temperature } T_{2}} \\ {\text { (iii) Expanded isothermally at } T_{2} \text { to state } d .} \\ {\text { (iv) Cooled at constant volume back to the initial state } a \text { . }}\end{array}\end{equation} Assume that the working fluid is \(n\) moles of an ideal gas (for which \(C_{V}\) is independent of temperature). (a) Calculate \(Q, W,\) and \(\Delta U\) for each of the processes \(a \rightarrow b, b \rightarrow c, c \rightarrow d,\) and \(d \rightarrow a\) . (b) In the Stirling cycle, the heat transfers in the processes \(b \rightarrow c\) and \(d \rightarrow a\) do not involve external heat sources but rather use regeneration: The same substance that transfers heat to the gas inside the cylinder in the process \(b \rightarrow c\) also absorbs heat back from the gas in the process \(d \rightarrow a .\) Hence the heat transfers \(Q_{b \rightarrow c}\) and \(Q_{d \rightarrow a}\) do not play a role in determining the efficiency of the engine. Explain this last statement by comparing the expressions for \(Q_{b \rightarrow c}\) and \(Q_{d \rightarrow a}\) calculated in part (a). (c) Calculate the efficiency of a Stirlingcycle engine in terms of the temperatures \(T_{1}\) and \(T_{2} .\) How does this compare to the efficiency of a Carnot-cycle engine operating between these same two temperatures? (Historically, the Stirling cycle was devised before the Carnot cycle.) Does this result violate the second law of thermodynamics? Explain. Unfortunately, actual Stirling-cycle engines cannot achieve this efficiency due to problems with the heat-transfer processes and pressure losses in the engine.

A Gasoline Engine. A gasoline 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 \(4.60 \times 10^{4} \mathrm{J} / \mathrm{g} .(\mathrm{a})\) 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?

As a budding mechanical engineer, you are called upon to design a Carnot engine that has 2.00 mol of a monatomic ideal gas as its working substance and operates from a hightemperature reservoir at \(500^{\circ} \mathrm{C}\) . The engine is to lift a 15.0 -kg weight 2.00 \(\mathrm{m}\) per cycle, using 500 \(\mathrm{J}\) of heat input. The gas in the engine chamber can have a minimum volume of 5.00 \(\mathrm{L}\) during the cycle. (a) Draw a \(p V\) -diagram for this cycle. Show in your diagram where heat enters and leaves the gas. (b) What must be the temperature of the cold reservoir? (c) What is the thermal efficiency of the engine? (d) How much heat energy does this engine waste per cycle? (e) What is the maximum pressure that the gas chamber will have to withstand?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.