Heat Engine: Second Law of Thermodynamics Numerical on Refrigerator Class-11 Nootan Physics Solutions Ch-21. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Heat Engine: Second Law of Thermodynamics Numerical on Refrigerator
( Class-11 Nootan Physics Solutions Ch-21 )
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Book | Nootan |
| Chapter-21 | Heat Engine: Second Law of Thermodynamics |
| Topics | Numerical on Refrigerator |
| Academic Session | 2024-2025 |
Heat Engine: Second Law of Thermodynamics Numerical on Refrigerator
Definition: Any device for removing heat from a cold body and adding it to hotter place is called “Refrigerator” or a heat pump. It is based on the principle of second law of thermodynamics.
Que-11: A refrigerator has to transfer an average of 263 J of heat per second from temperature -10°C to 25°C. Calculate the average power consumed, assuming no energy losses in the process.
Ans: T2 / (T1 -T2) = Q2 / (Q1 – Q2)
=> (273 – 10) / 35 = 263 / (Q1 – 263)
=> 263/35 = 263 / (Q1 – 263)
=> Q1 = 263 + 35 = 298 J
∴ W = Q1 – Q2 = 35 J
∴ Power = W/t = 35/1 = 35 watt
Que-12: Refrigerator A works between -10°C and 27°C, while refrigerator B works between -27°C and 17°C, both removing heat equal to 2000 J from the freezer. Which of the two is better refrigerator?
Ans: First refrigerator
β1 = T2 / (T1 – T2) = (273 – 10) / 37 = 263 / 37 = 7.12
Second refrigerator
β2 = (273 – 27) / 44 = 246 / 44 = 5.59
refrigerator A is better.
Que-13: In a refrigerator, heat from inside at 270 K is transferred to a room at 300 K. How many calories of heat will be delivered to the room for each joule of electrical energy consumed ideally?
Ans: β1 = T2 / (T1 – T2) = 270 / (300 – 270) = 270 / 30 = 9
again, β = Q2 / W
=> 9 = Q2 / 1
=> Q2 = 9 Joule
=> 9 / 4.2 = 2.38 calorie
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