Question

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

491 J of heat is delivered, power input required is 209 W, and COP is 5.46.

Step-by-step solution

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.

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.

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 \( Q_H \):\[ Q_H = Q_C \cdot \frac{T_H}{T_C} = 415 \cdot \frac{320}{270} \]

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.

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:\[ P = W \times \text{cycles per second} = 76 \times \frac{165}{60} \text{ J/s} \approx 209 \text{ W} \]

Determine Coefficient of Performance (COP)

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

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.

Key concepts

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.