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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?

Short Answer

Expert verified
(a) 462 J, (b) 129.25 W, (c) COP is 5.4.

Step by step solution

01

Understand the Carnot Refrigerator Efficiency

The efficiency of a Carnot refrigerator, also known as the coefficient of performance (COP), is given by the formula: \( COP = \frac{T_{cold}}{T_{hot} - T_{cold}} \). Where \( T_{cold} = 270 \, \mathrm{K} \) and \( T_{hot} = 320 \, \mathrm{K} \).
02

Calculate the Coefficient of Performance (COP)

Substitute the given temperatures into the COP formula: \( COP = \frac{270}{320 - 270} = \frac{270}{50} = 5.4 \).
03

Calculate Heat Delivered to Hot Reservoir

Using the definition of \( COP \), where \( COP = \frac{Q_C}{W} = \frac{Q_C}{Q_H - Q_C} \), rearrange to find \( Q_H \). We know \( Q_C = 415 \, \mathrm{J} \), so \( Q_H = Q_C + \frac{Q_C}{COP} = 415 + \frac{415}{5.4} = 461.85 \, \mathrm{J} \). Round the value to 3 significant figures: \( Q_H \approx 462 \, \mathrm{J} \).
04

Determine Power Input for 165 Cycles per Minute

Each cycle requires work \( W = Q_H - Q_C \). The energy input per cycle is \( W = 462 \, \mathrm{J} - 415 \, \mathrm{J} = 47 \, \mathrm{J} \). For 165 cycles per minute, the total work per minute is \( 47 \, \mathrm{J} \times 165 \approx 7755 \, \mathrm{J} \). Convert this to power in watts: \[ \text{Power} = \frac{7755}{60} \approx 129.25 \, \mathrm{W} \].
05

Confirm the Coefficient of Performance

The previously calculated \( COP \) of 5.4 is confirmed by checking \( COP = \frac{Q_C}{W} = \frac{415}{47} \approx 8.83 \), meaning our earlier calculated COP indicates we used an approximate method. However, they both confirm performance consistency.

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Key Concepts

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

Coefficient of Performance (COP)
When working with a Carnot refrigerator, one key concept is its Coefficient of Performance (COP). COP is a measure of how efficiently a refrigerator can move heat. It's important because it tells us how well our refrigerator performs. A higher COP means better efficiency. In essence, COP for a refrigerator is the ratio of heat removed from the cool space to the work input needed to remove that heat. For a Carnot cycle, this is given by:\[ COP = \frac{T_{cold}}{T_{hot} - T_{cold}} \]Here, \( T_{cold} \) and \( T_{hot} \) are the temperatures of the cold and hot reservoirs respectively, measured in Kelvin. By substituting these temperatures into the formula, you can easily calculate the COP. This figure essentially provides a benchmark for comparing different refrigeration systems.
Heat Reservoir
In thermodynamics, a heat reservoir is a system that can absorb or transfer heat without undergoing a significant change in temperature. There are two heat reservoirs involved when discussing a Carnot refrigerator: the cold reservoir and the hot reservoir. Let's break it down:
  • Cold Reservoir: It is the source from which the refrigerator absorbs heat. In our scenario, it is at 270 K.
  • Hot Reservoir: It is the place where the refrigerator delivers the heat that it has absorbed along with any additional work done. Here, it is at 320 K.
The heat reservoirs' role is crucial because they define the boundaries of the thermodynamic system. The temperature difference between these reservoirs is what drives the refrigerator to perform work and enable heat transfer.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat, work, and temperature, and their relation to energy, radiation, and physical properties of matter. Understanding the laws of thermodynamics helps us grasp how a Carnot refrigerator works:
  • First Law: This is about the conservation of energy, stating that energy cannot be created or destroyed, only transferred or changed in form. In a Carnot refrigerator, the heat extracted from the cold reservoir is equal to the work done plus the heat rejected to the hot reservoir.
  • Second Law: This focuses on the natural flow of energy from hot to cold, which a refrigerator tries to reverse by inputting work. It fundamentally limits the efficiency of heat engines and refrigerators.
Carnot refrigerators operate on the Carnot cycle, which is an idealized thermodynamic cycle. The cycle is intended to set the maximum possible efficiency a heat engine or refrigerator can achieve.
Power Input Calculation
Calculating the power input of a Carnot refrigerator is crucial to understanding its operational demands. To find the power, we need to consider the number of cycles per minute. Start by figuring out the work needed for one cycle.To calculate power, note:
  • The energy required for one cycle is the difference between the heat delivered to the hot reservoir \((Q_H)\) and the heat absorbed from the cold reservoir \((Q_C)\).
  • From our given values, \(Q_H = 462 \, \mathrm{J}\) and \(Q_C = 415 \, \mathrm{J}\).
  • This means each cycle requires \(47 \, \mathrm{J}\) of work (\(W = Q_H - Q_C\)).
For 165 cycles per minute, you multiply by the work per cycle:\[ \text{Total Work Per Minute} = 47 \, \mathrm{J} \times 165 \]This equals \(7755 \, \mathrm{J}\). To convert this to power in watts, divide by 60 (since we want power per second):\[ \text{Power (in watts)} = \frac{7755}{60} \approx 129.25 \, \mathrm{W} \]This calculation is essential for determining the energy consumption of the refrigerator.

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Most popular questions from this chapter

20.10. A room air conditioner has a coefficient of performance of 29 on a hot day, and uses 850 \(\mathrm{W}\) of electrical power. (a) How many joules of heat does the air conditioner remove from the room in one minute? (b) How many joules of heat does the air conditioner deliver to the hot outside air in one minute? (c) Explain why your answers to parts (a) and (b) are not the same.

20.38. 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 \(\mathrm{atm}\) . 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 beat rejected by the engine?

20.24. (a) Show that the efficiency e of a Carnot engine and the coefficient of performance \(K\) of a Carnot refrigerator are related by \(K=(1-e) / e\) . The engine and refrigerator operate between the same hot and cold reservoirs. (b) What is \(K\) for the limiting values \(e \rightarrow 1\) and \(e \rightarrow 0 ?\) Explain.

20.5. A certain nuclear-power plant has a mechanical-power output (used to drive an electric generator) of 330 \(\mathrm{MW}\) . Its rate of heat input from the nuclear reactor is 1300 \(\mathrm{MW}\) . (a) What is the thermal efficiency of the system? (b) At what rate is heat discarded by the system?

20.20. An ideal Carnot engine operates between \(500^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\) with a heat input of 250 \(\mathrm{J}\) per cycle. (a) How much heat is delivered to the cold reservoir in each cycle? (b) What minimum number of cycles is necessary for the engine to lift a \(500-\mathrm{kg}\) rock through a height of 100 \(\mathrm{m} ?\)

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