3000 J of heat is added to a system and 2500 J of work is done by the system. What is the change in internal energy of the system (in Joule)?

Question

BlackTom AI · Accepted Answer

The first law of thermodynamics gives ΔU = Q − W, where Q is the heat added to the system and W is the work done by the system. This sign convention (Q positive when heat enters the system, W positive when the system does work on the surroundings) is the relevant concept.

Step-by-step reasoning:
1) Identify quantities and signs:
   - Heat added to system: Q = +3000 J (positive because heat enters).
   - Work done by the system: W = 2500 J (positive by the chosen convention).
2) Apply the first law formula:
   ΔU = Q − W.
3) Substitute values and compute:
   ΔU = 3000 J − 2500 J
   ΔU = (3000 − 2500) J
   ΔU = 500 J.

Intermediate steps shown explicitly: subtract the work from the heat input: 3000 minus 2000 would be 1000, but here subtract 2500 giving 500. Units remain joules.

Why this answer is correct:
- Because the system receives 3000 J of energy as heat but expends 2500 J doing work on the surroundings, the net energy retained (internal energy increase) is the difference, 500 J. The algebra and sign conventions are consistent with standard thermodynamic notation.

Why other interpretations are incorrect:
- A common error is to use ΔU = Q + W; that form is used if W is defined as work done on the system (so-called alternate sign convention). If one mistakenly treated the 2500 J as work on the system, one would get 5500 J, which is wrong given the problem statement that the system does the work. Another mistake would be to take the result negative (−500 J), which would imply the system lost energy overall; but here heat in exceeds work out, so internal energy increases.

Conclusion: The change in internal energy is +500 J, which matches the provided answer 500.

3000 J of heat is added to a system and 2500 J of work is done by the system. What is the change in internal energy of the system (in Joule)?数值题

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