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A refrigerator is to be maintained at -73 degree Celsius and the outside air it as 27 degree Celsius. The minimum amount of work that must be supplied to remove 1000J of heat from inside the refrigerator is?
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A refrigerator is to be maintained at -73 degree Celsius and the outsi...
Analysis:
To solve this problem, we can use the concept of the Carnot cycle. The Carnot cycle is a theoretical thermodynamic cycle that consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.
Step 1: Identify the given values:
- The temperature inside the refrigerator (T1) = -73 °C = 200 K
- The temperature of the outside air (T2) = 27 °C = 300 K
- The amount of heat to be removed (Q) = 1000 J
Step 2: Calculate the efficiency of the Carnot cycle:
The efficiency (η) of the Carnot cycle is given by the formula:
η = 1 - (T1 / T2)
Substituting the given values into the formula:
η = 1 - (200 / 300)
= 1 - 2/3
= 1/3
Therefore, the efficiency of the Carnot cycle is 1/3 or 33.33%.
Step 3: Calculate the work done:
The work done (W) can be calculated using the formula:
W = Q / η
Substituting the given values into the formula:
W = 1000 / (1/3)
= 3000 J
Therefore, the minimum amount of work that must be supplied to remove 1000 J of heat from inside the refrigerator is 3000 J.
Conclusion:
To remove 1000 J of heat from inside the refrigerator, a minimum amount of 3000 J of work must be supplied. This can be achieved by using a Carnot cycle with an efficiency of 1/3. The Carnot cycle consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.
To solve this problem, we can use the concept of the Carnot cycle. The Carnot cycle is a theoretical thermodynamic cycle that consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.
Step 1: Identify the given values:
- The temperature inside the refrigerator (T1) = -73 °C = 200 K
- The temperature of the outside air (T2) = 27 °C = 300 K
- The amount of heat to be removed (Q) = 1000 J
Step 2: Calculate the efficiency of the Carnot cycle:
The efficiency (η) of the Carnot cycle is given by the formula:
η = 1 - (T1 / T2)
Substituting the given values into the formula:
η = 1 - (200 / 300)
= 1 - 2/3
= 1/3
Therefore, the efficiency of the Carnot cycle is 1/3 or 33.33%.
Step 3: Calculate the work done:
The work done (W) can be calculated using the formula:
W = Q / η
Substituting the given values into the formula:
W = 1000 / (1/3)
= 3000 J
Therefore, the minimum amount of work that must be supplied to remove 1000 J of heat from inside the refrigerator is 3000 J.
Conclusion:
To remove 1000 J of heat from inside the refrigerator, a minimum amount of 3000 J of work must be supplied. This can be achieved by using a Carnot cycle with an efficiency of 1/3. The Carnot cycle consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.
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A refrigerator is to be maintained at -73 degree Celsius and the outside air it as 27 degree Celsius. The minimum amount of work that must be supplied to remove 1000J of heat from inside the refrigerator is?
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A refrigerator is to be maintained at -73 degree Celsius and the outside air it as 27 degree Celsius. The minimum amount of work that must be supplied to remove 1000J of heat from inside the refrigerator is? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A refrigerator is to be maintained at -73 degree Celsius and the outside air it as 27 degree Celsius. The minimum amount of work that must be supplied to remove 1000J of heat from inside the refrigerator is? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A refrigerator is to be maintained at -73 degree Celsius and the outside air it as 27 degree Celsius. The minimum amount of work that must be supplied to remove 1000J of heat from inside the refrigerator is?.
A refrigerator is to be maintained at -73 degree Celsius and the outside air it as 27 degree Celsius. The minimum amount of work that must be supplied to remove 1000J of heat from inside the refrigerator is? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A refrigerator is to be maintained at -73 degree Celsius and the outside air it as 27 degree Celsius. The minimum amount of work that must be supplied to remove 1000J of heat from inside the refrigerator is? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A refrigerator is to be maintained at -73 degree Celsius and the outside air it as 27 degree Celsius. The minimum amount of work that must be supplied to remove 1000J of heat from inside the refrigerator is?.
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