| ⇦ | 
| ⇨ |
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: - The acceleration due to gravity becomes (g/2) where g=acceleration due to gravity
- A circuit area 0.01 m² is kept inside a magnetic field which is normal to its plane.
- A capacitor is charged to 200 volt. It has a charge of 0.1 coulomb.
- In a pure capacitive A.C circuit, current and voltage differ in phase by
- A stone is tied to a string of length l and whirled in a vertical circle with the other end
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The acceleration due to gravity becomes (g/2) where g=acceleration due to gravity
- A circuit area 0.01 m² is kept inside a magnetic field which is normal to its plane.
- A capacitor is charged to 200 volt. It has a charge of 0.1 coulomb.
- In a pure capacitive A.C circuit, current and voltage differ in phase by
- A stone is tied to a string of length l and whirled in a vertical circle with the other end
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