Chapter 20: Problem 17
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
Step by step solution
Understanding the Problem
Calculating the Work Done by the Refrigerator
Using Carnot Efficiency to Find Q_H
Compute Q_H
Calculate Power Input Required
Determine Coefficient of Performance (COP)
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Heat Reservoirs
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.
Thermodynamic Cycles
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.
Coefficient of Performance
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
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.