SATHEE: Thermodynamics Question 168

Thermodynamics Question 168

Question: An ideal refrigerator has a freezer at a temperature of $- 13^{\circ} C .$ The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) will be

[BHU 2000; CPMT 2002]

Options:

A) 325°C

B) 325K

C) 39°C

D) 320°C

Show Answer

Answer:

Correct Answer: C

Solution:

Coefficient of performance $K = \frac{T_{2}}{T_{1} - T_{2}}$

Therefore $5 = \frac{(273 - 13)}{T_{1} - (273 - 13)} = \frac{260}{T_{1} - 260}$

Therefore $5 T_{1} - 1300 = 260$

Therefore $5 T_{1} = 1560$

Therefore $T_{1} = 312 K \to 39^{\circ} C$

Thermodynamics Question 168

Question: An ideal refrigerator has a freezer at a temperature of $- 13^{\circ} C .$ The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) will be

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Thermodynamics Question 168

Question: An ideal refrigerator has a freezer at a temperature of −13∘C. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) will be

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: An ideal refrigerator has a freezer at a temperature of $- 13^{\circ} C .$ The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) will be