A certain refrigerator is rated as being \(32.0 \%\) as ef ficient as a Carnot refrigerator. To remove \(100 .\) J of heat from the interior at \(0^{\circ} \mathrm{C}\) and eject it to the outside at \(22^{\circ} \mathrm{C}\), how much work must the refrigerator motor do?

Short Answer

Expert verified
Answer: The work done by the refrigerator motor is approximately 4184.10 J.

Step by step solution

01

Convert the temperatures to Kelvin

We need to work with temperatures in Kelvin. To convert the given temperatures from Celsius to Kelvin, we will use the equation: \(K = °C + 273.15\) For the interior temperature (\(T_{cold}\)): \(T_{cold} = 0°C + 273.15 = 273.15 K\) For the exterior temperature (\(T_{hot}\)): \(T_{hot} = 22°C + 273.15 = 295.15 K\)
02

Calculate the efficiency of the Carnot refrigerator

The efficiency of a Carnot refrigerator (or heat engine) is given by the equation: \(Efficiency_{Carnot} = 1 - \frac{T_{cold}}{T_{hot}}\) Substitute the calculated temperatures from Step 1: \(Efficiency_{Carnot} = 1 - \frac{273.15}{295.15} \approx 0.0746\)
03

Calculate the efficiency of the given refrigerator

We know that the given refrigerator is \(32.0 \%\) as efficient as a Carnot refrigerator. To find the efficiency of the given refrigerator, we will multiply the efficiency of the Carnot refrigerator by \(0.32\). \(Efficiency_{given} = 0.32 \times Efficiency_{Carnot} = 0.32 \times 0.0746 \approx 0.0239\)
04

Calculate the work done by the refrigerator motor

We are given that the refrigerator removes \(100 J\) of heat from the interior. Using the efficiency of the given refrigerator, we can find the work done by the motor with the equation: \(Work = \frac{Q_{removed}}{Efficiency_{given}}\) Substitute the given heat removed (\(100 J\)) and the calculated efficiency of the given refrigerator: \(Work = \frac{100}{0.0239} \approx 4184.10 J\) So, the work done by the refrigerator motor is approximately \(4184.10 J\).

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.

Thermodynamic Temperature Conversion
Understanding the conversion between different temperature scales is crucial when working with thermodynamic systems, as these often require temperatures to be expressed in an absolute scale such as Kelvin. The Kelvin scale is an absolute temperature scale, starting at absolute zero, the theoretical point where all molecular motion ceases.

To convert Celsius to Kelvin, the operation is straightforward: you simply add 273.15 to the Celsius temperature. This is key in thermodynamics because calculations involving gas laws, heat cycles, or any thermodynamic process rely on absolute temperatures to ensure accuracy.

For example, if we have a temperature of 20°C and we want to convert it to Kelvin, the equation would look like this:
\(K = 20°C + 273.15 = 293.15 K\). This process of temperature conversion is foundational when analyzing refrigeration systems like the Carnot refrigerator.
Carnot Cycle Efficiency
The Carnot cycle efficiency is vital in understanding the theoretical limits of a heat engine or refrigerator’s performance. For a Carnot refrigerator, the efficiency is defined by the difference in temperature between the cold reservoir (inside of the fridge) and the hot reservoir (the environment), both expressed in Kelvin.

The Carnot efficiency equation is:
\(Efficiency_{Carnot} = 1 - \frac{T_{cold}}{T_{hot}}\).

This formula allows us to understand that as the temperature difference between the two reservoirs decreases, the Carnot refrigerator becomes more efficient. However, actual refrigerators can never be 100% efficient due to practical constraints and the Second Law of Thermodynamics.
Refrigeration Work Calculation
When calculating the work required for a refrigeration process, it is essential to understand the concept of efficiency. For a given refrigerator, we express efficiency as a fraction of the Carnot efficiency. To calculate the work, \(W\), done by the refrigerator motor, one must divide the amount of heat, \(Q_{removed}\), extracted from the cold reservoir by the efficiency of the refrigerator, \(Efficiency_{given}\):

\(Work = \frac{Q_{removed}}{Efficiency_{given}}\).

This formula demonstrates how the refrigerator’s work relates directly to its efficiency. Lower efficiency requires more work to remove the same amount of heat. In practice, this corresponds to more significant electrical energy consumption, essential for environmental and economic reasons.
Kelvin Temperature Scale
The Kelvin scale is an absolute temperature scale used predominantly in the scientific community because it is directly related to the energy of particles. Zero Kelvin, or absolute zero, is the point where particles have the least possible energy and are not moving. Unlike Celsius or Fahrenheit, Kelvin does not use 'degrees'.

The Kelvin temperature scale is imperative in thermodynamics, including studies of heat engines and refrigerators, as calculations depend on an absolute temperature measurement. The conversion from Celsius to Kelvin helps students properly calculate heat transfer, work, and efficiency in thermodynamic systems.

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

Assume that it takes \(0.0700 \mathrm{~J}\) of energy to heat a \(1.00-\mathrm{g}\) sample of mercury from \(10.000^{\circ} \mathrm{C}\) to \(10.500{ }^{\circ} \mathrm{C}\) and that the heat capacity of mercury is constant, with a negligible change in volume as a function of temperature. Find the change in entropy if this sample is heated from \(10 .{ }^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\).

Find the net change in entropy when \(100 . \mathrm{g}\) of water at \(0^{\circ} \mathrm{C}\) is added to \(100 . \mathrm{g}\) of water at \(100 .{ }^{\circ} \mathrm{C}\)

A refrigerator with a coefficient of performance of 3.80 is used to \(\operatorname{cool} 2.00 \mathrm{~L}\) of mineral water from room temperature \(\left(25.0^{\circ} \mathrm{C}\right)\) to \(4.00^{\circ} \mathrm{C} .\) If the refrigerator uses \(480 . \mathrm{W}\) how long will it take the water to reach \(4.00^{\circ} \mathrm{C}\) ? Recall that the heat capacity of water is \(4.19 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K}),\) and the density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\). Assume the other contents of the refrig. erator are already at \(4.00^{\circ} \mathrm{C}\).

The number of macrostates that can result from rolling a set of \(N\) six-sided dice is the number of different totals that can be obtained by adding the pips on the \(N\) faces that end up on top. The number of macrostates is a) \(6^{N}\) b) \(6 N\) c) \(6 N-1\). d) \(5 N+1\).

A coal-burning power plant produces \(3000 .\) MW of thermal energy, which is used to boil water and produce supersaturated steam at \(300 .{ }^{\circ} \mathrm{C}\). This high-pressure steam turns a turbine producing \(1000 .\) MW of electrical power. At the end of the process, the steam is cooled to \(30.0^{\circ} \mathrm{C}\) and recycled. a) What is the maximum possible efficiency of the plant? b) What is the actual efficiency of the plant? c) To cool the steam, river water runs through a condenser at a rate of \(4.00 \cdot 10^{7} \mathrm{gal} / \mathrm{h}\). If the water enters the condenser at \(20.0^{\circ} \mathrm{C}\), what is its exit temperature?

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