| ⇦ | 
| ⇨ |
A Carnot engine, having an efficiency of η=1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature is
Options
(a) 100 J
(b) 99 J
(c) 90 J
(d) 1 J
Correct Answer:
90 J
Explanation:
Efficiency of carnot engine n = 1 – (T₂ / T₁)
That is, 1 / 10 = 1 – (T₂ / T₁)
⇒ (T₂ / T₁) = 1 – (1 / 10) = 9 / 10 ⇒ (T₁ / T₂) = 10 / 9
.·. w = Q₂ . [(T₁ / T₂) – 1) ⇒ 10 = Q₂ [(10 / 9) – 1]
⇒ 10 = Q₂ (1 / 9) ⇒ Q₂ = 90 J
So, 90 J heat is absorbed at lower temperature.
Related Questions: - In the adiabatic compression, the decrease in volume associated with
- When radiation is incident on a photoelectron emitter,
- The shortest wavelength in Lyman series is 91.2 nm. The largest wavelength
- An adiabatic process occurs at constant
- If the nucleus ²⁷₁₃Al has nuclear radius of about 3.6 fm
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In the adiabatic compression, the decrease in volume associated with
- When radiation is incident on a photoelectron emitter,
- The shortest wavelength in Lyman series is 91.2 nm. The largest wavelength
- An adiabatic process occurs at constant
- If the nucleus ²⁷₁₃Al has nuclear radius of about 3.6 fm
Topics: Thermodynamics (179)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply