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A person does 30 kJ work on 2 kg of wate...

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?

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Work done on the system (by the person while stirring), `W =- 30 kJ = - 30,000 J`
Heat flowing out of the system, `Q = -" 5 kcal"= -5 xx 4184 J = 20920 J`
Using First law of thermodynamics `Delta U=Q-W`
`Delta U=-20,920J-(30,000)J`
`DeltaU=-20,920J+30,000J=9080J`
Here, the heat lost is than the work done on the system, so that change in internal energy is positive.
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