1.5 kJ of heat was absorbed by a system, whilst also having 7.5 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

The First Law of Thermodynamics states ΔU = Q + W, where Q is heat added to the system and W is work done on the system. Because the problem explicitly says heat was absorbed and work was done on the system, both Q and W are positive.

1) Relevant knowledge points and concepts
- First Law: ΔU (change in internal energy) = Q + W.
- Sign convention used here: Q > 0 when heat is absorbed by the system; W > 0 when work is done on the system.

2) Step-by-step reasoning
- Identify Q: "1.5 kJ of heat was absorbed" ⇒ Q = +1.5 kJ.
- Identify W: "7.5 kJ of work done on it" ⇒ W = +7.5 kJ.
- Apply the First Law: ΔU = Q + W.

3) Calculations with intermediate steps
- ΔU = 1.5 kJ + 7.5 kJ
- ΔU = (1.5 + 7.5) kJ
- ΔU = 9.0 kJ

4) Why the correct answer is correct
- Both contributions increase the system’s internal energy: heat added raises energy and work done on the system also raises energy. Summing them per the law gives +9.0 kJ.

5) Why other possible answers would be incorrect
- If one mistakenly uses the sign for work done by the system (W positive when done by the system) they would set W = −7.5 kJ and get ΔU = 1.5 − 7.5 = −6.0 kJ; that is incorrect because the statement specified work was done on the system, not by it.
- Another common error is subtracting the smaller from the larger without regard to sign convention (yielding 6.0 kJ); this ignores that both processes here add energy.

6) Conclusion confirming the final answer
- Therefore the change in internal energy is ΔU = 9.0 kJ, which matches the provided answer.

1.5 kJ of heat was absorbed by a system, whilst also having 7.5 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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