Question

Water is being heated in a closed pan on top of a range while being stirred by a paddle wheel. During the process, 30 kJ of heat is transferred to the water, and 5 kJ of heat is lost to the surrounding air. The paddle-wheel work amounts to 500 N ∙ m. Determine the final energy of the system if its initial energy is 10 kJ. [

Answer 1

Answer:

35.5KJ

Explanation:

Heat transferred (Q) = change in internal energy (∆U) + work done (W)

Q = 30KJ, W = 500Nm = 500J = 500/1000KJ = 0.5KJ

∆U = Q - W = 30 - 0.5 = 25.5KJ

∆U = final energy (U2) - initial energy (U1)

U2 = ∆U + U1 = 25.5 + 10 = 35.5KJ

Answer 2

The heat transfer of a closed system is the addition of change in internal energy and the total amount of work done by it.

The final energy of the system is 35.5 kJ.

What is heat transfer?

The heat transfer is the transfer of thermal energy due to the temperature difference.

The heat transfer of a closed system is the addition of change in internal energy and the total amount of work done by it.

Thus the heat transferred of a system can be given as,

$Q=\Delta U +W$

Here, $\Delta U$ is the change in internal energy and $W$ is the amount of work done.

Given information-

The paddle-wheel work amounts to 500 N-m or 0.5 kJ.

The initial energy of the system is 10 kJ.

Total heat transferred during the process is 30 kJ.

Total heat lost during the process to the surrounding air is 5 kJ.

Put the values in the above formula as,

$30=\Delta U +0.5$

As the change in internal energy is the sum of final energy $U_f$ and the initial energy $U_i$ of the system. Thus,

$30=U_f+U_i +0.5\\30=U_f+10+0.5\\U_f=35.5\rm kJ$

Hence the final energy of the system is 35.5 kJ.

Learn more about the heat transfer here;

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