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A Carnot refrigerator transfers heat from its inside \(\left(6.0^{\circ} \mathrm{C}\right)\) to the room air outside \(\left(20.0^{\circ} \mathrm{C}\right)\). (a) Find the coefficient of performance of the refrigerator. (b) Determine the magnitude of the minimum work needed to \(\operatorname{cool} 5.00 \mathrm{~kg}\) of water from 20.0 to \(6.0^{\circ} \mathrm{C}\) when it is placed in the refrigerator.

Short Answer

Expert verified
COP: 19.94; Minimum Work: 14.70 kJ

Step by step solution

01

Convert Temperatures to Kelvin

The first step is to convert the given temperatures from Celsius to Kelvin.The conversion formula is:\[ T(K) = T(°C) + 273.15 \]For the inside temperature: \( T_\text{inside} = 6.0 + 273.15 = 279.15 \text{ K} \)For the outside temperature: \( T_\text{outside} = 20.0 + 273.15 = 293.15 \text{ K} \)
02

Calculate the Coefficient of Performance

The coefficient of performance (COP) for a Carnot refrigerator is given by the formula:\[ \text{COP} = \frac{T_C}{T_H - T_C} \]where \( T_C \) is the inside (cold) temperature and \( T_H \) is the outside (hot) temperature, both in Kelvin.Substitute the values:\[ \text{COP} = \frac{279.15}{293.15 - 279.15} = \frac{279.15}{14} \approx 19.94 \]
03

Calculate Heat Removal Energy

The heat removed (\( Q_C \)) from the water can be calculated using the specific heat formula:\[ Q_C = m \cdot c \cdot \Delta T \]where \( m = 5.00 \text{ kg} \) is the mass of the water, \( c = 4.186 \text{ kJ/kg°C} \) is the specific heat capacity of water, and \( \Delta T = 20.0 - 6.0 = 14.0 \text{ °C} \) is the change in temperature.Substitute the values:\[ Q_C = 5.00 \times 4.186 \times 14 = 293.02 \text{ kJ} \]
04

Determine Minimum Work (a)

The minimum work done by the refrigerator is calculated using the relation between work, COP, and heat removed:\[ W = \frac{Q_C}{\text{COP}} \]Substitute the known values:\[ W = \frac{293.02 \text{ kJ}}{19.94} \approx 14.70 \text{ kJ} \]
05

Conclusion

The coefficient of performance of the refrigerator is \( 19.94 \). The minimum work needed to cool \( 5.00 \text{ kg} \) of water from \( 20.0 \text{°C} \) to \( 6.0 \text{°C} \) is approximately \( 14.70 \text{ kJ} \).

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Key Concepts

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

Coefficient of Performance
The Coefficient of Performance (COP) is a key measure in evaluating the efficiency of refrigerators. Unlike efficiency in traditional engines, which is a ratio of useful output to input, the COP is calculated as the ratio of the heat removed from the cold space to the work input. For a Carnot refrigerator, it is expressed as:\[ \text{COP} = \frac{T_C}{T_H – T_C} \]where:- \( T_C \) is the temperature inside the refrigerator (cold reservoir temperature) in Kelvin.- \( T_H \) is the temperature of the room air outside the refrigerator (hot reservoir temperature) in Kelvin.The higher the COP, the more efficient the refrigerator is at transferring heat for the same amount of work input. It's essential for understanding how much work is saved compared to the heat extracted. For example, a COP of 19.94 means the system provides about 19.94 units of heat removal for every unit of work input.
Thermodynamics
Thermodynamics is a branch of physics that deals with heat, work, and the forms of energy involved in physical and chemical processes. When studying refrigerators, we explore how these devices transfer heat from a low-temperature area to a higher temperature one against the natural flow, which would normally move from hot to cold. This process appears somewhat contrary to our everyday experience. Key principles include: - **First Law of Thermodynamics:** Energy cannot be created or destroyed, only transformed. In a refrigerator, the work input is converted into heat transfer. - **Second Law of Thermodynamics:** In any cycle, the sum of the transformations is equal to the heat extracted, which means perfect energy conversion is impossible. Hence, efficiencies are never 100%, and this is why COPs are used to gauge performance.
Specific Heat Capacity
The concept of specific heat capacity is crucial when calculating the energy required to change the temperature of a substance. Specific heat capacity (\(c\)) is defined as the amount of heat required to change the temperature of one kilogram of a substance by one degree Celsius or Kelvin. For water, this is \(4.186 \, \text{kJ/kg°C}\), meaning it's relatively high compared to other substances.This property is significant in thermodynamics when calculating the heat exchange, as seen in the formula:\[ Q_C = m \cdot c \cdot \Delta T \] where:- \( m \) is the mass of the substance- \( c \) is the specific heat capacity- \( \Delta T \) is the change in temperatureFor example, the amount of energy required to cool the water in our scenario is calculated using this formula, highlighting the large energy input needed to change water temperature.
Kelvin Temperature Conversion
Temperature conversion to the Kelvin scale is essential when performing calculations in thermodynamics, particularly for problems involving the Carnot cycle. Kelvin is the absolute thermodynamic temperature scale starting from absolute zero, where molecules have minimal kinetic energy.Conversion formula:\[ T(\text{K}) = T(\degree C) + 273.15 \] This formula is used to convert temperatures into Kelvin, as thermodynamic equations typically require absolute temperature for accuracy. Absolute temperature enables a consistent reflection of thermal energy, unaffected by the negative or positive connotations of Celsius or Fahrenheit scales.Understanding temperature in Kelvin helps avoid potential errors in calculations like the COP, where exact differences between temperatures in Kelvin can influence results significantly.

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Most popular questions from this chapter

A Carnot engine has an efficiency of 0.700 , and the temperature of its cold reservoir is \(378 \mathrm{~K}\). (a) Determine the temperature of its hot reservoir. (b) If \(5230 \mathrm{~J}\) of heat is rejected to the cold reservoir, what amount of heat is put into the engine?

A fifteen-watt heater is used to heat a monatomic ideal gas at a constant pressure of \(7.60 \times 10^{5} \mathrm{~Pa}\) During the process, the \(1.40 \times 10^{-3} \mathrm{~m}^{3}\) volume of the gas increases by \(25.0 \%\). How long was the heater on?

Due to a tune-up, the efficiency of an automobile engine increases by \(5.0 \% .\) For an input heat of \(1300 \mathrm{~J},\) how much more work does the engine produce after the tune-up than before?

A system gains a certain amount of energy in the form of heat at constant pressure, and the internal energy of the system increases by an even greater amount. (a) Is any work done and, if so, is it done on or by the system? (b) If there is work, is it positive or negative, according to our convention? (c) Does the volume of the system increase, decrease, or remain the same? Give your reasoning A system gains \(2780 \mathrm{~J}\) of heat at a constant pressure of \(1.26 \times 10^{5} \mathrm{~Pa}\) and its internal energy increases by \(3990 \mathrm{~J}\). What is the change in volume of the system, and is it an increase or a decrease? Verify that your answer is consistent with your answers to the Concept Questions.

Three moles of an ideal gas are compressed from \(5.5 \times 10^{-2}\) to \(2.5 \times 10^{-2} \mathrm{~m}^{3}\) During the compression, \(6.1 \times 10^{3} \mathrm{~J}\) of work is done on the gas, and heat is removed to keep the temperature of the gas constant at all times. Find (a) \(\Delta U,\) (b) \(Q,\) and (c) the temperature of the gas.

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