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Example · Example 2

Q.A system absorbs 500 J500\ \text{J} of heat from its surroundings and simultaneously does 300 J300\ \text{J} of work on the surroundings by expanding. Calculate the change in internal energy, ΔU\Delta U, of the system.

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The system absorbs heat, so q=+500 Jq = +500\ \text{J} (positive, by the IUPAC convention). The system does 300 J300\ \text{J} of work ON the surroundings (it expands), so the work done ON the system is w=−300 Jw = -300\ \text{J} (negative). Applying the First Law: ΔU=q+w=500+(−300)=200 J\Delta U = q + w = 500 + (-300) = 200\ \text{J}. The internal energy of the system increases by 200 J200\ \text{J} — the heat absorbed was more than enough to supply the work done in expanding, with the surplus increasing the system's internal energy. [!ANSWER] ΔU=+200 J\Delta U = +200\ \text{J}.

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