| ⇦ |
| ⇨ |
The internal energy change in a system that has absorbed 2 kcals of heat and done 500J of work is:
Options
(a) 6400 J
(b) 5400 J
(c) 7900 J
(d) 8900 J
Correct Answer:
7900 J
Explanation:
According to first law of thermodynamics Q = ∆U + W
∆U = Q – W = 2 x 4.2 x 1000 – 500 = 8400 – 500 = 7900 J
Related Questions: - The average power dissipated in a pure inductor is
- A battery having e.m.f. 4 volt and internal resistance 0.5 Ω is connected
- If a spring of spring constant k stretched by a length x under tension T, the energy
- Photocells convert
- The kinetic energy of particle moving along a circle of radius R depends
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The average power dissipated in a pure inductor is
- A battery having e.m.f. 4 volt and internal resistance 0.5 Ω is connected
- If a spring of spring constant k stretched by a length x under tension T, the energy
- Photocells convert
- The kinetic energy of particle moving along a circle of radius R depends
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