In a certain process, 200J of heat energy is supplied to a system and at the same time 50J of work is done by the system. The increase in the internal energy of the system is:
Correct answer: C. 150 J
- A. 25 J
- B. 100 J
- C. 150 J
- D. 250 J
Explanation
The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W). In this case:ΔU = Q - WGiven that 200J of heat energy is supplied to the system (Q = 200J) and 50J of work is done by the system (W = 50J), we can substitute these values into the equation:ΔU = 200J - 50J = 150JThus, the increase in internal energy is 150J, making Option C the correct choice.Options A, B, and D are incorrect because they either misapply the first law of thermodynamics or ignore the work done by the system, leading to inaccurate calculations of internal energy change.
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About First Law of Thermodynamics
The first law relates heat supplied, work done and the change in internal energy through energy conservation. Problems use sign conventions and apply the law to isothermal, adiabatic, isobaric and isochoric processes. Internal energy is a state function, while heat and work depend on the path followed.
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