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Questions 4-12 · Q5

Q.A system releases 130 kJ of heat while 109 kJ of work is done on the system. Calculate the change in internal energy.

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✓ Free question

Step 1. The system RELEASES 130 kJ of heat, so by the sign convention (heat leaving the system is negative), Q=−130Q = -130 kJ.

Step 2. 109 kJ of work is done ON the system, so the work done BY the system is W=−109W = -109 kJ.

Step 3. By the First Law, ΔU=Q−W=(−130)−(−109)=−130+109=−21\Delta U = Q - W = (-130) - (-109) = -130 + 109 = -21 kJ.

Step 4. So the system's internal energy DECREASES by 21 kJ — the textbook's printed answer, "ΔU = 21 kJ", gives the correct magnitude but omits the sign; since more heat leaves the system (130 kJ) than work is put into it (109 kJ), the net effect must be a decrease in internal energy, i.e. ΔU = −21 kJ, not +21 kJ.

✓Final answer

ΔU = −21 kJ (a decrease of 21 kJ — the textbook's printed "21 kJ" gives only the magnitude; since the system loses more heat than the work put into it, ΔU must be negative)

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