Entropy due to this irreversible process

[Simobartz]

Homework Statement

There is an adiabatic container filled with water. The volume of the container is fixed and inside the continer there is a propeller electrically connected to the outside. If the work done by the propeller on the water is $\delta W_{prop}$, what is the entropy generated due to irreversibilities?

Relevant Equations

$$dU=TdS-PdV$$

I think the solution is:
$$dU=\delta W_{prop}$$
$$dU=TdS-PdV$$
$$dV=0$$
then, $$TdS=\delta W_{prop}$$ and so $$dS=dU/T$$
and by the way, it correct to say that, if the transformation between the initial and the final state would happen in a reversible way then the heat transfer could be calculated as
$$dS=\delta Q_{rev} /T$$
$$\delta Q _{rev}=TdS$$
$$\delta Q_{rev}=\delta W_{prop}$$

 

[Andrew Mason]

Homework Statement:: There is an adiabatic container filled with water. The volume of the container is fixed and inside the container there is a propeller electrically connected to the outside. If the work done by the propeller on the water is $\delta W_{prop}$, what is the entropy generated due to irreversibilities?
Relevant Equations:: $$dU=TdS-PdV$$

I think the solution is:
$$dU=\delta W_{prop}$$
$$dU=TdS-PdV$$
$$dV=0$$
then, $$TdS=\delta W_{prop}$$ and so $$dS=dU/T$$
and by the way, it correct to say that, if the transformation between the initial and the final state would happen in a reversible way then the heat transfer could be calculated as
$$dS=\delta Q_{rev} /T$$
$$\delta Q _{rev}=TdS$$
$$\delta Q_{rev}=\delta W_{prop}$$

You have the right idea. Since entropy is a state function, entropy change does not depend on the process in going from the initial state to the final state. So it is a matter of finding a convenient reversible path from initial to final state and calculating $\Delta S = \int_{i}^{f}\frac{dQ_{rev}}{T}$ along that path.

The equivalent reversible path would be heat flow into the water increasing internal energy by an amount equal to the work done.

Since W=Q=mC(Tf-Ti), [where m is the mass of the water and C is its specific heat capacity], $T_f=T_i+W/mC$ and $dW = dQ = mCdT$.

$\Delta S = \int_{T_i}^{T_i+W/mC}\frac{dQ_{rev}}{T} = \int_{T_i}^{T_i+W/mC}\frac{mCdT}{T}$
So:
$\Delta S = mC\ln\left(\frac{T_i+W/mC}{T_i}\right)$

AM