NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1
The internal energy change in a system that has absorbed 2 kcal of heat and done 500 J of work is
- A6400 J
- B5400 J
- C7900 J
- D8900 J
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Correct answer: C
When a quantity Q of heat is supplied to a system it is used to do an amount of work W by the system and to increase the internal energy of the system by ∆U :
Q = ∆U + W
Here, Q given in kilo calories is converted into joule.
Therefore Q = 2×1000×4.2 J = 8400 J
The increase the internal energy
∆U = Q – W = 8400 – 500 = 7900 J
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