⇦ | ⇨ |
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: - A body is thrown vertically upward in air when air resistance is taken into account
- A photocell employs photoelectric effect to convert
- A metal bar of length L and area of cross-section A is clamped between two rigid
- If 75% of the radioactive reaction is completed in 2 hrs, what would be
- Wires A and B are made from the same material. A has twice the diameter and three times
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body is thrown vertically upward in air when air resistance is taken into account
- A photocell employs photoelectric effect to convert
- A metal bar of length L and area of cross-section A is clamped between two rigid
- If 75% of the radioactive reaction is completed in 2 hrs, what would be
- Wires A and B are made from the same material. A has twice the diameter and three times
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