Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Expert verified
(a) 1.48 × 10^4 J, (b) 4.58 × 10^4 J.

Step by step solution

01

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.
02

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 \( W \).
03

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.
04

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 \( 1.48 \times 10^4 \text{ J} \).
05

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 \( Q_h = Q_c + W \).
06

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 \( 4.58 \times 10^4 \text{ J} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A Carnot engine is operated between two heat reservoirs at temperatures of 520 \(K\) and 300 \(K\). (a) If the engine receives 6.45 \(kJ\) of heat energy from the reservoir at 520 \(K\) in each cycle, how many joules per cycle does it discard to the reservoir at 300 \(K\)? (b) How much mechanical work is performed by the engine during each cycle? (c) What is the thermal efficiency of the engine?

A cylinder contains oxygen at a pressure of 2.00 atm. The volume is 4.00 \(L\), and the temperature is 300 \(K\). Assume that the oxygen may be treated as an ideal gas. The oxygen is carried through the following processes: (i) Heated at constant pressure from the initial state (state 1) to state 2, which has \(T\) = 450 K. (ii) Cooled at constant volume to 250 \(K\) (state 3). (iii) Compressed at constant temperature to a volume of 4.00 \(L\) (state 4). (iv) Heated at constant volume to 300 \(K\), which takes the system back to state 1. (a) Show these four processes in a \(pV\)-diagram, giving the numerical values of \(p\) and \(V\) in each of the four states. (b) Calculate \(Q\) and \(W\) for each of the four processes. (c) Calculate the net work done by the oxygen in the complete cycle. (d) What is the efficiency of this device as a heat engine? How does this compare to the efficiency of a Carnot cycle engine operating between the same minimum and maximum temperatures of 250 \(K\) and 450 \(K\)?

A box is separated by a partition into two parts of equal volume. The left side of the box contains 500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature. The partition is punctured, and equilibrium is eventually attained. Assume that the volume of the box is large enough for each gas to undergo a free expansion and not change temperature. (a) On average, how many molecules of each type will there be in either half of the box? (b) What is the change in entropy of the system when the partition is punctured? (c) What is the probability that the molecules will be found in the same distribution as they were before the partition was punctured- that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right half?

Digesting fat produces 9.3 food calories per gram of fat, and typically 80% of this energy goes to heat when metabolized. (One food calorie is 1000 calories and therefore equals 4186 J.) The body then moves all this heat to the surface by a combination of thermal conductivity and motion of the blood. The internal temperature of the body (where digestion occurs) is normally 37\(^\circ\)C, and the surface is usually about 30\(^\circ\)C. By how much do the digestion and metabolism of a 2.50-g pat of butter change your body's entropy? Does it increase or decrease?

Compare the entropy change of the warmer water to that of the colder water during one cycle of the heat engine, assuming an ideal Carnot cycle. (a) The entropy does not change during one cycle in either case. (b) The entropy of both increases, but the entropy of the colder water increases by more because its initial temperature is lower. (c) The entropy of the warmer water decreases by more than the entropy of the colder water increases, because some of the heat removed from the warmer water goes to the work done by the engine. (d) The entropy of the warmer water decreases by the same amount that the entropy of the colder water increases.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free