Does work have the same impact on entropy change as heat?

[Est120]

1. Homework Statement
if a rigid adiabatic container has a fan inside that provides 15000 j of work to an ideal gas inside the container,
does the change in entropy would be the same as if 15000 j of heat are provided to the same rigid container (removing the insulation)?2. Relevant equations
rigid means that the volume of the container is constant ,so is a contant volume process

3. The attempt at a solution
i managed to do the calculations and got the right answer(that is why i don't give so many facts),my doubt is does work changes entropy the same amount as heat does?

 

[stockzahn]

In an idealised (reversible) work is/can be provided isentropically (so no entropy increase) - for turbomachinery (to which a fan more less is counted to) the ideal isentropic process is taken to validate the quality of the real process. In a real process, due to losses, part of the work will be transformed into heat and the entropy will increase.

However, the fan provides work to the air and accelerates it, therefore part of the work supplied is stored as kinetic energy in the air. The energy stored as kinetic energy does not contribute to an entropy increase, therefore an amount of energy supplied as heat increases the entropy more than the same amount of energy provided as work. With time due to friction, viscosity and other losses the air will be decelerated and all the kinetic energy is transoformed into heat, resulting in the same rise entropy.

So heat and work do not cause the same entropy change in the first place, only after a certain amount of time, if the work/kinetic energy isn't extracted, friction will transform it into heat, yielding the same final entropy.

Edit: In your case the change of state happens isochorally, that means the process is done along a line of constant volume. With decreasing speed of the air, the state is shifted to higher temperatures and higher entropies on the isochor.

 

[Chestermiller]

In your example, the heat and the work will result in the same entropy change because entropy depends only on the thermodynamic state of the system (and not how that state was achieved), and, in your example, the initial and final thermodynamic equilibrium states of the system are the same for the work case as for the heat case. To get the entropy change (at least for the work case), what you need to do is devise a reversible path between the initial and final thermodynamic equilibrium states and calculate the integral of dq/T for that path.