Maximum Work from Carnot Engine

[ts547]

Homework Statement

A 200 litre container of boiling water and a 200 litre container of ice cold water are used
as heat source and sink for a Carnot engine. Calculate the maximum amount of useful
work that can be obtained from the system and the final temperature of the two
containers of water.

Homework Equations

W=Q_h-Q_c
W=(1-T_c/T_h)*Q_h

Q=nC(T2-T1)?

The Attempt at a Solution

Im getting mega confused. I thought the carnot engine was a cyclic process i.e the temperatures of the resevoirs don't change. For example the heat energy transfers are during isothermal expansions in the reservoirs so I get

W=nR(T_h-T_c)ln(V_2/V_1)

but we don't know one of the V's. Does a 200 litre container imply that it's full? In that case the heat transfer would have to come from an isochoric process as the container is a fixed volume. That would make sense in order to find the temperature change but that's the second part of the Q. Otherwise the max work out would be when the log term is max but that is infinity.

Can you find Q_h from knowing just the temperatures and volumes of the reservoirs? That would solve all my problems. Well, its a start...

Thanks for your help in advance

 

[Mapes]

The Carnot engine is indeed cyclic, and the temperature (but not the volume) of the reservoirs changes by an infinitesimal amount during each cycle. Your job is to find the extracted work and final temperature after a very, very large (in theory, infinite) number of cycles.

Try writing the change in entropy, the change in energy, and the change in temperature of each reservoir during a typical cycle, assuming constant volume and constant heat capacity. Then try to link these equations together, knowing that a Carnot cycle is reversible.

 

[Andrew Mason]

If you make the engine small enough you can get arbitrarily close to a reversible isothermal/adiabatic path for each cycle.

To calculate the work done directly is a difficult task. There is an easier way to do it using entropy. Take Mape's suggestion and work out the expression for change in entropy after the reservoirs have reached the same temperature, T. With that, you should be able to determine T and from that the work done.

AM

 

[ts547]

Ok so I've got,

Change in entropy of hot reservoir:

ds_h=dQ/T_h=> deltaS_h = C(T-T_h)/T_h

where T is the final equilibrium temperature, Q is for constant volume process

Change in entropy of cold reservoir:

deltaS_c = C(T-T_c)/T_c

and change in entropy of the engine system:

deltaS_system = Cln(Th/Tc)

add them together and set equal to zero for reversible process, also make one of the reservoirs negative (hot reservoir by convention i think) and I get:

T=317.97K

which is between 0 and 100 degrees so looks good :)

so max work out is:

W=Q_h(1-T_c/T_h)

where

Q_h=C(T-T_h) ? (sort of a question, usually told in the question to approximate C as 3R/2 if we need to use it)

and I get:

W=-184J (work out is negative by convention in thermo? I always forget)

 

[Mapes]

Not quite. The temperatures of the reservoirs are changing slowly with time, so you can't use the original reservoir temperatures in the denominators of the entropy equations. Try writing in differential form and then integrating over the temperature change.

 

[Andrew Mason]

...

and change in entropy of the engine system:

deltaS_system = Cln(Th/Tc)

There is no material change in entropy of the system. Technically, the system does not quite return to its intial state after each cycle, since the temperature of the hot reservoir is continually declining. But, if you make the engine infintessimally small, the change in entropy of the system can be made arbitrarily small so as to be ignored.

Take Mapes' suggestion and integrate dQh/T over the range Th to T (hint: what is dQh in terms of dT?). Do the same for the cold reservoir.

AM

 

[ts547]

Ok try this,

as dv=0, dU=dQ, dU=CdT=dQ,

dS=CdT/T for both reservoirs with relavant limits,

deltaS=Cln(T/Th)+Cln(T/Tc)=0

=> T/Th=Tc/T

=> T=sqrt(ThTc)=319.26K

I presume my max work theory was ok.

Also while your here, I am getting confused again. I don't know wots wrong with me. Must be christmas.

The next part is,

If the containers were connected by a conducting bar so that they came into thermal
equilibrium calculate the final temperature and the change in entropy of the system.

Is it an appropriate assumption that there is the same mass of water at 100C as there is at 0C in the same volume. So then the equilibrium temperature is half way between. Seems to simple. We know that the change in entropy is zero at equilibrium but then Th=Tc which doesn't help with equations above. Thank You.

 

[Mapes]

Just integrate your dS term again. Of course, now it's not equal to zero.