2.1 kJ of heat was released by a system, whilst also having 9.2 kJ of work done on it. What is the change in the internal energy of this system after these processes?

Question

BlackTom AI · Accepted Answer

Apply the first law of thermodynamics in the sign convention where Q is heat added to the system and W is work done on the system: ΔU = Q + W.

Relevant concepts:
- Internal energy change ΔU is equal to net energy transferred as heat plus net energy transferred as work.
- Sign convention matters: heat released by the system is negative Q; work done on the system is positive W (under ΔU = Q + W). Equivalently, if using ΔU = Q − W (where W is work done by the system), convert signs consistently.

Step-by-step reasoning and calculations:
1. Heat: "2.1 kJ of heat was released by a system" means Q = −2.1 kJ (energy leaves the system).
2. Work: "9.2 kJ of work done on it" means W = +9.2 kJ (energy enters the system as work).
3. Use ΔU = Q + W:
   ΔU = (−2.1 kJ) + (+9.2 kJ)
4. Compute the sum:
   −2.1 + 9.2 = 7.1
   So ΔU = +7.1 kJ.

Why this is correct:
- The positive result means the system’s internal energy increases by 7.1 kJ because the work done on the system (9.2 kJ in) exceeds the heat lost (2.1 kJ out) by 7.1 kJ.
- If using the alternate formula ΔU = Q − W_by (where W_by is work done by the system), note W_by = −9.2 kJ here, so ΔU = −2.1 − (−9.2) = 7.1 kJ — same result.

Common incorrect answers and why they’re wrong:
- −7.1 kJ would correspond to wrong sign for the work (treating work on the system as negative).
- −11.3 kJ or +11.3 kJ come from adding magnitudes without correct signs (2.1 + 9.2 = 11.3) but ignoring that heat was released.
- Confusion arises only from inconsistent sign conventions; consistent use leads to +7.1 kJ.

Conclusion: The change in internal energy is +7.1 kJ, which matches the provided answer.

2.1 kJ of heat was released by a system, whilst also having 9.2 kJ of work done on it. What is the change in the internal energy of this system after these processes?数值题

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