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A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 K and 270 K. (a) If in each cycle the refrigerator receives 415 J of heat energy from the reservoir at 270 K, how many joules of heat energy does it deliver to the reservoir at 320 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
491 J of heat is delivered, power input required is 209 W, and COP is 5.46.

Step by step solution

01

Understanding the Problem

We have a Carnot refrigerator operating between two reservoirs at 320 K and 270 K. For each cycle, it receives 415 J from the cooler (270 K) reservoir. We need to calculate how much heat is delivered to the warmer (320 K) reservoir.
02

Calculating the Work Done by the Refrigerator

For a Carnot refrigerator, the relationship between the absorbed heat, delivered heat, and work done is given by: \[ Q_H = Q_C + W \] where \( Q_C = 415 \text{ J} \) is the heat absorbed from the cold reservoir, \( Q_H \) is the heat delivered to the hot reservoir, and \( W \) is the work input for each cycle.
03

Using Carnot Efficiency to Find Q_H

The efficiency of a Carnot refrigerator can be expressed as:\[ \text{Efficiency} = \frac{T_H - T_C}{T_H} \]We rearrange the efficiency formula to get\[ \frac{Q_H - Q_C}{Q_H} = \frac{T_H - T_C}{T_H} \]Substitute \( T_H = 320 \text{ K} \) and \( T_C = 270 \text{ K} \):\[ \frac{T_H}{T_C} = \frac{Q_H}{Q_C} \]Solve for \( Q_H \):\[ Q_H = Q_C \cdot \frac{T_H}{T_C} = 415 \cdot \frac{320}{270} \]
04

Compute Q_H

Substitute the values into the equation:\[ Q_H = 415 \times \frac{320}{270} \approx 491 \text{ J} \]Thus, the refrigerator delivers approximately 491 J of heat to the reservoir at 320 K.
05

Calculate Power Input Required

The power input required is calculated using work per cycle and cycles per minute. We know the number of cycles per minute is 165, so:\[ W = Q_H - Q_C = 491 - 415 = 76 \text{ J} \]Power is given by:\[ P = W \times \text{cycles per second} = 76 \times \frac{165}{60} \text{ J/s} \approx 209 \text{ W} \]
06

Determine Coefficient of Performance (COP)

The Coefficient of Performance for a refrigerator is given by:\[ \text{COP} = \frac{Q_C}{W} \]Calculate \( COP \) using:\[ \text{COP} = \frac{415}{76} \approx 5.46 \]
07

Conclusion

For the Carnot refrigerator, \( Q_H \approx 491 \text{ J} \), the power input is approximately 209 W, and the COP is approximately 5.46.

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Key Concepts

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

Heat Reservoirs
In the context of a Carnot refrigerator, heat reservoirs play a central role. A heat reservoir is an idealized body large enough to absorb or supply finite amounts of heat without undergoing any temperature change. In this exercise, two different heat reservoirs are involved: a cold reservoir at 270 K and a hot reservoir at 320 K.

The cold reservoir supplies heat to the refrigerator, while the hot reservoir receives the heat expelled by the refrigerator. This heat exchange is crucial for the operation of the refrigerator.
  • The cold reservoir is essential as it provides the necessary heat input that the refrigerator needs to extract.
  • The hot reservoir is where this extracted heat is delivered, often leading to useful heating applications, or simply discharging excess heat.
Understanding these heat flow and reservoir interactions is vital for appreciating how a Carnot refrigerator operates effectively and efficiently.
Thermodynamic Cycles
Thermodynamic cycles are sequences of processes that return a system to its initial state, allowing the continuous operation of devices like refrigerators. The Carnot cycle is a specific type of thermodynamic cycle used in refrigerators and heat engines for achieving maximum efficiency.

The Carnot cycle consists of four stages: two isothermal (constant temperature) processes and two adiabatic (no heat exchange) processes. These stages allow the refrigerator to absorb and expel heat efficiently. In this exercise, the Carnot refrigerator undergoes cycles between the two reservoirs at different temperatures.
  • During the isothermal expansion at the lower temperature, the refrigerator absorbs heat from the cold reservoir, doing work.
  • During isothermal compression at the higher temperature, it releases the absorbed heat to the hot reservoir.
This cycle repeats, enabling the refrigerator to maintain temperature difference between two reservoirs and keep the inside of the fridge at a lower temperature.
Coefficient of Performance
The Coefficient of Performance (COP) is an important measure of a refrigerator's efficiency. It indicates how effectively a refrigerator removes heat from the low-temperature reservoir and how much work (or energy) is used to do so.

COP is defined as the ratio of the heat extracted from the cold reservoir (Q_C) to the work input (W).\[\text{COP} = \frac{Q_C}{W}\]In our exercise, the heat taken from the cold reservoir is 415 J, and the work done per cycle is 76 J. Hence, the COP is calculated as:

\[\text{COP} = \frac{415}{76} \approx 5.46\]A higher COP value means a more efficient refrigerator, as it indicates more heat transfer per unit of work input. It's a critical parameter for assessing the effectiveness of cooling devices.
Carnot Efficiency
Carnot efficiency is a theoretical maximum efficiency that any heat engine or refrigerator can achieve when operating between two temperatures. This concept provides a yardstick against which the performance of real-life refrigerators and engines can be measured.

For a Carnot refrigerator, the efficiency is determined by the temperature difference between the hot (T_H) and cold (T_C) reservoirs.
  • The formula for Carnot efficiency is given by:\[\eta = 1 - \frac{T_C}{T_H} \]
  • This exercise involves reservoirs at 270 K and 320 K, allowing us to calculate the maximum theoretical efficiency.
The efficiency calculated here serves as a theoretical limit; no real refrigerator can exceed it due to unavoidable energy losses in practical systems. Understanding this limitation helps engineers design better refrigerators that approach this ideal efficiency, paving the way for more energy-efficient technologies.

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Most popular questions from this chapter

A person with skin of surface area 1.85 m\(^2\) and temperature 30.0\(^\circ\)C is resting in an insulated room where the ambient air temperature is 20.0\(^\circ\)C. In this state, a person gets rid of excess heat by radiation. By how much does the person change the entropy of the air in this room each second? (Recall that the room radiates back into the person and that the emissivity of the skin is 1.00.)

A 15.0-kg block of ice at 0.0\(^\circ\)C melts to liquid water at 0.0\(^\circ\)C inside a large room at 20.0\(^\circ\)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).

You decide to take a nice hot bath but discover that your thoughtless roommate has used up most of the hot water. You fill the tub with 195 kg of 30.0\(^\circ\)C water and attempt to warm it further by pouring in 5.00 kg of boiling water from the stove. (a) Is this a reversible or an irreversible process? Use physical reasoning to explain. (b) Calculate the final temperature of the bath water. (c) Calculate the net change in entropy of the system (bath water + boiling water), assuming no heat exchange with the air or the tub itself.

A certain brand of freezer is advertised to use 730 kW \(\cdot\) h of energy per year. (a) Assuming the freezer operates for 5 hours each day, how much power does it require while operating? (b) If the freezer keeps its interior at -5.0\(^\circ\)C in a 20.0\(^\circ\)C room, what is its theoretical maximum performance coefficient? (c) What is the theoretical maximum amount of ice this freezer could make in an hour, starting with water at 20.0\(^\circ\)C?

An air conditioner operates on 800 W of power and has a performance coefficient of 2.80 with a room temperature of 21.0\(^\circ\)C and an outside temperature of 35.0\(^\circ\)C. (a) Calculate the rate of heat removal for this unit. (b) Calculate the rate at which heat is discharged to the outside air. (c) Calculate the total entropy change in the room if the air conditioner runs for 1 hour. Calculate the total entropy change in the outside air for the same time period. (d) What is the net change in entropy for the system (room + outside air) ?

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