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The inside of a Carnot refrigerator is maintained at a temperature of 277 \(\mathrm{K}\) , while the temperature in the kitchen is 299 \(\mathrm{K}\) . Using 2500 \(\mathrm{J}\) of work, how much heat can this refrigerator remove from its inside compartment?

Short Answer

Expert verified
The refrigerator can remove 31,475 J of heat from its inside compartment.

Step by step solution

01

Understand the Carnot Refrigerator

A Carnot refrigerator is a type of idealized heat engine that transfers heat from a lower temperature reservoir to a higher temperature one, using work input. The efficiency of a Carnot refrigerator is dependent on the temperatures of these reservoirs.
02

Apply the Coefficient of Performance (COP) formula

The Coefficient of Performance (COP) for a Carnot refrigerator is given by \( \text{COP} = \frac{T_c}{T_h - T_c} \), where \( T_c \) is the cold reservoir temperature (inside compartment) and \( T_h \) is the hot reservoir temperature (kitchen temperature).
03

Calculate the COP

Substitute the given temperatures into the COP formula: \( T_c = 277 \, \mathrm{K} \) and \( T_h = 299 \mathrm{K} \). Thus, \( \text{COP} = \frac{277}{299 - 277} = \frac{277}{22} \approx 12.59 \).
04

Compute Heat Removed

The heat removed (\( Q_c \)) from the inside of the refrigerator can be calculated using \( Q_c = \text{COP} \times W \), where \( W \) is the work done on the refrigerator (2500 J). Substituting the calculated COP, \( Q_c = 12.59 \times 2500 = 31475 \, \mathrm{J} \).

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Key Concepts

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

Coefficient of Performance (COP)
When we talk about the efficiency of refrigerators, the Coefficient of Performance (COP) is a key metric. It measures how effectively a refrigerator can transfer heat from a cold area (like inside the fridge) to a warmer area by using external work input.
For refrigerators, a higher COP means better performance, as it indicates more heat removal with less work. COP is crucial because it directly relates to energy efficiency, an important aspect in both economic and environmental terms.
In the Carnot cycle, the COP is determined using the formula \( \text{COP} = \frac{T_c}{T_h - T_c} \). This means the performance depends on the temperature difference between the heat reservoirs. The larger the temperature gap, the harder and less efficient it becomes to move heat.
thermodynamics
Thermodynamics is the study of energy transfer and conversions. It applies to all systems that involve matter and energy exchange. In refrigerators, thermodynamics explains the process of cooling by understanding how heat moves from one place to another.
One of the fundamental principles is the Second Law of Thermodynamics, which tells us that heat will naturally flow from a warmer area to a cooler one, unless work is done. Thus, refrigerators work against this natural flow by using work input to transfer heat from inside the fridge to the warmer outside. This operation relies on cycles like the Carnot cycle, which represents an idealized process that maximizes efficiency.
heat engine
A heat engine is any device that converts thermal energy into work. In the context of a refrigerator, it works in reverse to a typical heat engine: it uses work (like electricity) to transfer heat, instead of producing work from heat.
The concept of a heat engine is central to understanding how refrigerators can cool down the inside of their compartments. They draw heat from the interior and release it into the external environment. The Carnot cycle represents an ideal heat engine, maximizing efficiency based on the temperature difference between the inside and outside of the fridge.
temperature reservoirs
Temperature reservoirs play a fundamental role in the function of heat engines and refrigerators. They are sections at uniform temperature, either absorbing or emitting heat.
  • Cold Reservoir: This is the area from which the refrigerator absorbs heat. In the exercise, it is the inside of the refrigerator at 277 K.
  • Hot Reservoir: Heat is expelled here, typically the surrounding kitchen environment at 299 K.
Understanding temperature reservoirs helps us grasp the energy dynamics within refrigeration. The difference in temperature between these reservoirs defines the efficiency, as shown in the COP formula. Lower temperature differences result in higher efficiencies, making it easier and less energy-consuming to remove heat.

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Most popular questions from this chapter

Heat is added to two identical samples of a monatomic ideal gas. In the first sample the heat is added while the volume of the gas is kept constant, and the heat causes the temperature to rise by 75 \(\mathrm{K}\) . In the second sample, an identical amount of heat is added while the pressure (but not the volume) of the gas is kept constant. By how much does the temperature of this sample increase?

One mole of neon, a monatomic gas, starts out at conditions of standard temperature and pressure. The gas is heated at constant volume until its pressure is tripled, then further heated at constant pressure until its volume is doubled. Assume that neon behaves as an ideal gas. For the entire process, find the heat added to the gas.

A Carnot engine operates with a large hot reservoir and a much smaller cold reservoir. As a result, the temperature of the hot reservoir remains constant while the temperature of the cold reservoir slowly increases. This temperature change decreases the efficiency of the engine to 0.70 from \(0.75 .\) Find the ratio of the final temperature of the cold reservoir to its initial temperature.

A Carnot refrigerator is used in a kitchen in which the temperature is kept at 301 \(\mathrm{K}\) . This refrigerator uses 241 \(\mathrm{J}\) of work to remove 2561 \(\mathrm{J}\) of heat from the food inside. What is the temperature inside the refrigerator?

A Carnot engine operates between temperatures of 650 and 350 \(\mathrm{K}\) . To improve the efficiency of the engine, it is decided either to raise the temperature of the hot reservoir by 40 \(\mathrm{K}\) or to lower the temperature of the cold reservoir by 40 \(\mathrm{K}\) . Which change gives the greater improvement? Justify your answer by calculating the efficiency in each case.

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