Question

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

(a) 1.48 × 10^4 J, (b) 4.58 × 10^4 J.

Step-by-step solution

Understanding Coefficient of Performance

The coefficient of performance (COP) is a measure of a refrigerator's efficiency and is defined by the formula \( \text{COP} = \frac{Q_c}{W} \), where \( Q_c \) is the heat absorbed from the cold reservoir and \( W \) is the work (mechanical energy) required.

Given Values

We know from the problem statement that \( \text{COP} = 2.10 \) and \( Q_c = 3.10 \times 10^4 \text{ J} \). These are the key values needed to find the work \( W \).

Calculate Mechanical Energy Required

Rearrange the COP formula to solve for \( W \): \( W = \frac{Q_c}{\text{COP}} \). Substitute the given values: \( W = \frac{3.10 \times 10^4}{2.10} \). Calculate \( W \) to find the mechanical energy.

Calculation for Part a

\( W = \frac{3.10 \times 10^4}{2.10} \approx 1.48 \times 10^4 \text{ J} \). The mechanical energy required each cycle is approximately \( 1.48 \times 10^4 \text{ J} \).

Calculate Heat Discarded to High-Temperature Reservoir

Use the energy conservation principle: the heat discarded \( Q_h \) to the high-temperature reservoir is given by \( Q_h = Q_c + W \).

Calculation for Part b

Substitute the known values to find \( Q_h \): \( Q_h = 3.10 \times 10^4 + 1.48 \times 10^4 = 4.58 \times 10^4 \text{ J} \). The heat discarded to the high-temperature reservoir is \( 4.58 \times 10^4 \text{ J} \).

Key concepts

Understanding Thermodynamics

Thermodynamics is a foundational concept in physics that deals with the movement and transformation of energy. It is crucial for understanding the principles behind heat engines, refrigerators, and many other systems.

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.

In the context of a refrigerator, these principles help explain how the system moves heat from a cooler area to a warmer one, requiring work (mechanical energy) to do so. The coefficient of performance (COP) is a measure of this efficiency.

Explaining Heat Transfer in Refrigerators

Heat transfer is the process of thermal energy moving from one place to another. It's essential in the operation of refrigerators where heat is transferred away from the interior.

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.

For efficient operation, understanding the relationship between these energy exchanges is essential. The heat rejected to the high-temperature reservoir is always larger than the heat absorbed from the cold one, due to added mechanical energy.

Efficiency Calculation and Coefficient of Performance

Efficiency in thermodynamic cycles, like those in refrigerators, is assessed using different metrics. For a refrigerator, the Coefficient of Performance (COP) is key.

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.

In our exercise, the provided COP and the heat absorbed from the cold reservoir allow us to calculate the work required (\(W\)) using the formula rearranged as:\[W = \frac{Q_c}{\text{COP}}\]

• 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.

Ultimately, efficient energy transfer saves energy and reduces costs, which is why calculating efficiency is so crucial in thermodynamics.