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NEET PHYSICSEasy

The internal energy change in a system that has absorbed 2 kcal2 \text{ kcal} of heat and done 500 J500 \text{ J} of work is:

A

8900 J

B

6400 J

C

5400 J

D

7900 J

Step-by-Step Solution

According to the First Law of Thermodynamics, the relationship between heat (QQ), work (WW), and the change in internal energy (ΔU\Delta U) is given by ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W (using the standard physics sign convention where work done by the system is positive).

  1. Convert Heat to Joules: Given heat absorbed Q=2 kcal=2000 calQ = 2 \text{ kcal} = 2000 \text{ cal}. Using the mechanical equivalent of heat (1 cal4.2 J1 \text{ cal} \approx 4.2 \text{ J}): Q=2000×4.2 J=8400 JQ = 2000 \times 4.2 \text{ J} = 8400 \text{ J}.

  2. Identify Work Done: Work done by the system W=500 JW = 500 \text{ J}.

  3. Calculate Internal Energy Change: Rearranging the First Law equation: ΔU=QW\Delta U = Q - W. ΔU=8400 J500 J=7900 J\Delta U = 8400 \text{ J} - 500 \text{ J} = 7900 \text{ J}.

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