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Q.A person does 30 kJ work on 2 kg of water by stirring using a paddle wheel. While stirring, around 5 kcal of heat is released from water through its container to the surface and surroundings by thermal conduction and radiation. What is the change in internal energy of the system ?

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2022Subjective· 3mImportance★★★★★
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Use the first law of thermodynamics, being careful with signs: work is done ON the water (by the paddle wheel, so it adds energy), while heat is LOST from the water (to the surroundings, so it removes energy). The net effect here is a small overall increase in internal energy.

First law of thermodynamics (in the convention ΔU = Q + W, where Q is heat ADDED to the system and W is work done ON the system):

ΔU = Q + W

Given:

Work done ON the water by the paddle wheel (stirring): W = +30 kJ (positive, since it is work done ON the system, adding energy to it).

Heat released FROM the water to the surroundings: 5 kcal. Converting to kJ (using 1 cal = 4.2 J, so 1 kcal = 4.2 kJ):

Heat released = 5 × 4.2 = 21 kJ

Since this heat LEAVES the system, Q = −21 kJ (negative, heat lost).

Substitute into the first law:

ΔU = Q + W = (−21 kJ) + (30 kJ) = +9 kJ …

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