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?

A person does 30 kJ work on 2 kg of water by stirring using a paddle w

1.	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?
Answer» Solution :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.