A system releases 10 kJ of heat and performs 15 kJ of work on the surrounding. Hence the change in internal energy is :

Q.1 (vi) A system releases 10 kJ of heat and performs 15 kJ of work on the surrounding. Hence the change in internal energy is :

A. + 5 kJ

B. – 5 kJ

C. + 25 kJ

D. – 25 kJ

Doubtly Asked 27 Oct 2026 Board exam · March 2026 hsc 12th Save WhatsApp Report

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Answer: D. – 25 kJ

Answer: (d) –25 kJ

First law: \(\Delta U = Q + W\). Heat released by the system: \(Q = -10\) kJ. Work done by the system on the surroundings: \(W = -15\) kJ. So \(\Delta U = -10 + (-15) = -25\) kJ.

Doubtly AI · 27 Oct 2026 AI-verified · Oct 2026 Report

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A system releases 10 kJ of heat and performs 15 kJ of work on the surrounding. Hence the change in internal energy is :

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