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20.18. A Carnot device extracts 5.00 \(\mathrm{kJ}\) of heat from a body at \(-10.0^{\circ} \mathrm{C} .\) How much work is done if the device exhausts heat into the environment at \((a) 25.0^{\circ} \mathrm{C} ;(\mathrm{b}) 0.0^{\circ} \mathrm{C} ;(\mathrm{c})-25.0^{\circ} \mathrm{C} ;\) In each case, is the device acting as an engine or as a refrigerator?

Short Answer

Expert verified
(a) Work: 585 J, Refrigerator; (b) Work: 185 J, Refrigerator; (c) No work, Non-functional.

Step by step solution

01

Convert Temperatures to Kelvin

The first step is to convert all given temperatures from Celsius to Kelvin using the formula: \( T(K) = T(°C) + 273.15 \). \(-10.0^{\circ} \text{C} = 263.15 \text{K}\), \(25.0^{\circ} \text{C} = 298.15 \text{K}\), \(0.0^{\circ} \text{C} = 273.15 \text{K}\), \(-25.0^{\circ} \text{C} = 248.15 \text{K}\).
02

Calculate Efficiency of Carnot Device

For a Carnot device, the efficiency can be calculated using the formula: \( \eta = 1 - \frac{T_c}{T_h} \), where \( T_h \) is the temperature of the hot reservoir and \( T_c \) is the temperature of the cold reservoir.
03

Case a: Efficiency and Work at 25.0°C

Here, \( T_h = 298.15 \text{K} \) and \( T_c = 263.15 \text{K} \). Calculate efficiency: \( \eta = 1 - \frac{263.15}{298.15} \approx 0.117 \). Work done: \( W = \eta \times Q_h = 0.117 \times 5000 \text{ J} \approx 585 \text{ J} \). The device acts as a refrigerator because heat is absorbed at a lower temperature and released at a higher temperature.
04

Case b: Efficiency and Work at 0.0°C

Here, \( T_h = 273.15 \text{K} \) and \( T_c = 263.15 \text{K} \). Efficiency: \( \eta = 1 - \frac{263.15}{273.15} \approx 0.037 \). Work done: \( W = \eta \times Q_h = 0.037 \times 5000 \text{ J} \approx 185 \text{ J} \). The device acts as a refrigerator because it extracts heat from a cold reservoir and exhausts it to a warmer environment.
05

Case c: Efficiency and Work at -25.0°C

Here, \( T_h = 248.15 \text{K} \) and \( T_c = 263.15 \text{K} \). Since the hot temperature is less than the cold temperature, the efficiency is negative: \( \eta = 1 - \frac{263.15}{248.15} \). This scenario isn't physically valid for a Carnot cycle acting as a refrigerator or engine, and thus no work is done and the device does not function.
06

Conclusion

In cases a and b, the device acts as a refrigerator due to the natural direction of heat flow (hot to cold). In case c, the device cannot function because the cold reservoir is warmer than the hot reservoir.

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with heat, work, and the interrelation of various forms of energy. It helps us understand how energy moves and changes form in various systems. In thermodynamics, we often talk about systems, which can be as small as a single gas container or as large as a power plant.
Some important concepts in thermodynamics include:
  • The First Law of Thermodynamics: Energy cannot be created or destroyed, only transferred or transformed.
  • The Second Law of Thermodynamics: Heat energy will naturally flow from a hot body to a cold one.
The Carnot cycle, which you encountered in this exercise, is an idealized sequence of processes that provides a standard by which we can measure real engines' efficiency. It's important because it defines the maximum possible efficiency a heat engine can achieve.
Understanding thermodynamics is crucial for designing efficient engines, refrigerators, and other systems that rely on heat transfer.
Heat Engine
A heat engine is a device that converts thermal energy into mechanical work. The engine absorbs heat from a high-temperature reservoir, performs work, and then releases some heat to a lower-temperature reservoir.
The Carnot engine is an idealized example of a heat engine where all processes are reversible and there are no energy losses. In your exercise, when the Carnot device operates as a heat engine, it would absorb heat (like the 5.00 kJ here), convert a portion into work, and exhaust the remainder to a cold reservoir.
Key points about a heat engine include:
  • It follows a cycle, continuously converting heat into work.
  • Its efficiency depends on the temperature difference between the hot and cold reservoirs.
The effectiveness of the conversion process is encapsulated by its efficiency, which is the ratio of work output to heat input.
Refrigerator
A refrigerator is essentially the reverse of a heat engine. Instead of converting heat into work, it uses work to move heat from a cold reservoir to a hot one.
In the context of the Carnot cycle, when the given device acts as a refrigerator, as in cases (a) and (b) of your exercise, it extracts heat from the colder environment and expels it to a warmer one. This is crucial in cooling applications, like domestic fridges and air conditioners.
Key characteristics of a refrigerator include:
  • It requires an external work input to move heat against the natural flow (from cold to hot).
  • Its performance is often measured by the Coefficient of Performance (CoP), which is different from efficiency.
A CoP quantifies how effectively a refrigerator performs its function compared to the energy input required.
Efficiency
Efficiency is a measure of how well a system transforms input energy into useful work. In a perfect Carnot cycle, efficiency is determined by the temperatures of the reservoirs it operates between. The formula for Carnot efficiency is given by:\[\eta = 1 - \frac{T_c}{T_h}\]Where:
  • \(T_c\) is the absolute temperature of the cold reservoir.
  • \(T_h\) is the absolute temperature of the hot reservoir.
In your exercise, this formula helped calculate the efficiency of the Carnot device in different scenarios. Efficiency is crucial because it tells us what fraction of heat intake can be converted into useful work.
Real-world systems always have efficiencies less than a Carnot engine due to losses and irreversibilities. However, understanding this theoretical limit helps engineers strive for designs that approach it, optimizing the balance of work done and energy used.

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Most popular questions from this chapter

20.17. A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 \(\mathrm{K}\) and 270 \(\mathrm{K}\) (a) If in each cycle the refrigerator receives 415 \(\mathrm{J}\) of heat energy from the reservoir at 270 \(\mathrm{K}\) , how many joules of heat energy does it deliver to the reservoir at 320 \(\mathrm{K} ?\) (b) If the refrigerator completes 165 cycles each minute, what power input is required to operate it? (c) What is the coefficient of performance of the refrigerator?

20.27 A 15.0-kg block of ice at \(0.0^{\circ} \mathrm{C}\) melts to liquid water at \(0.0^{\circ} \mathrm{C}\) inside a large room that has a temperature of \(20.0^{\circ} \mathrm{C}\) . Treat the ice and the room as an isolated system, and assume that the room is large enough for its temperature change to be ignored. (a) Is the melting of the ice reversible or irreversible? Explain, using simple physical reasoning without resorting to any equations. (b) Calculate the net entropy change of the system during this process. Explain whether or not this result is consistent with your answer to part (a).

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?

20.52. A typical coal-fired power plant generates 1000 MW of usable power at an overall thermal efficiency of 40\(\%\) (a) What is the rate of heat input to the plant? (b) The plant burns anthracite coal, which has a heat of combustion of \(265 \times 10^{7} \mathrm{J} / \mathrm{kg}\) . How much coal does the plant use per day, if it operates continuously? (c) At what rate is heat ejected into the cool reservoir, which is the nearby river?(d) The river's temperature is \(18.0^{\circ} \mathrm{C}\) before it reaches the power plant and \(18.5^{\circ} \mathrm{C}\) after it has received the plant's waste heat. Calculate the river's flow rate, in cubic meters per second. (e) By how much does the river's entropy increase each second?

20.9. A refrigerator has a coefficient of performance of \(2.10 .\) In cach cycle it absorbs \(3.40 \times 10^{4} \mathrm{J}\) of heat from the cold reservoir. (a) How much mechanical energy is required each cycle to operate the refrigerator? (b) During each cycle, how much heat is discarded to the high-temperature reservoir?

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