Question

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 1

Given,

$\longrightarrow\sf{\Delta W=-30\ kJ}$

(since work is done on the system.)

$\longrightarrow\sf{\Delta Q=-5\ kcal=-21\ kJ}$

(since heat is liberated from the system.)

Then by first law of thermodynamics,

$\longrightarrow\sf{\Delta u=\Delta Q-\Delta W}$

$\longrightarrow\sf{\Delta u=-21\ kJ+30\ kJ}$

$\longrightarrow\sf{\underline {\underline {\Delta u=9\ kJ}}}$

Answer 2

Answer:

Explanation:

Work done ==> W=−30kJ=−30,000J

Heat flow ==> Q=− 5 kcal=−5×4184J=20920J

We know ,

First law of thermodynamics :

ΔU=Q−W

ΔU=−20,920J−(30,000)J

ΔU=−20,920J+30,000J=9080J

ΔU = 9.080KJ

Hence , the change in internal energy of the system = 9.08 KJ

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