Chapter 20: Problem 9
A refrigerator has a coefficient of performance of 2.10. In each cycle it absorbs 3.10 \(\times\) 10\(^4\) 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?
Short Answer
Step by step solution
Understanding Coefficient of Performance
Given Values
Calculate Mechanical Energy Required
Calculation for Part a
Calculate Heat Discarded to High-Temperature Reservoir
Calculation for Part b
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Thermodynamics
Thermodynamics is built on the concept of energy conservation. In essence, energy cannot be created or destroyed, only transformed from one form to another. There are four fundamental laws of thermodynamics, but for this exercise, the first and second laws are most relevant.
- The First Law of Thermodynamics states that the total energy of an isolated system is constant, and energy can be transferred or converted but not destroyed.
- The Second Law of Thermodynamics introduces the concept of entropy, stating that energy transformations are not 100% efficient, and some energy is always dissipated as heat.
Explaining Heat Transfer in Refrigerators
Refrigerators work on the principle of extracting heat from a low-temperature area (inside the fridge) and transferring it to a high-temperature area (outside the fridge). This heat transfer requires an input of mechanical energy.
- A refrigerator absorbs heat from its contents, seen as the cold reservoir in thermodynamic terms.
- This absorbed heat (\(Q_c\)) is then expelled to the surroundings, defined as the high-temperature reservoir.
- The work (\(W\)) done by the refrigerator is derived from an external power source, usually electrical energy.
Efficiency Calculation and Coefficient of Performance
The COP of a refrigerator is a ratio that compares the heat removed from the cold reservoir to the work input required:\[\text{COP} = \frac{Q_c}{W}\]
- Higher COP values signify greater efficiency, meaning the refrigerator requires less work to remove a specific amount of heat.
- With a COP of 2.10 and \(Q_c = 3.10 \times 10^4 \text{ J}\), we find the work required is approximately \(1.48 \times 10^4 \text{ J}\).
- The efficiency encapsulated in COP highlights how much mechanical energy is utilized effectively to move heat.