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JEE PHYSICS MCQ: THERMODYNAMICS

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(A): QQ heat is given to melt ice cube and magnitude of work done by ice on surrounding is W.W. then ΔQ=Q−W\Delta Q=Q-W
(R): First law of thermodynamics is applicable in this situation.
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Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A)
Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because the first law of thermodynamics explains the relationship between heat, work, and internal energy change.

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