Ideal Carnot Engine: Find Time & Power Output

[vladimir69]

Homework Statement

An ideal Carnot engine operates between a heat reservoir and a block of ice of mass M. An external energy source maintains the reservoir at a constant temperature T_{h}. At time t=0 the ice is at its melting point T_{0}, but it is insulated from everything except the engine, so it is free to change state and temperature. The engine is operated in such a way that it extracts heat from the reservoir at a constant rate P_{h}.
(a) Find an expression for the time t_{1} at which all the ice is melted, in terms of the quantities given and any other thermodynamic parameters.
(b) Find an expression for the mechanical power output of the engine as a function of time for times t > t_{1}.

Homework Equations

Q=ML_{f}
W=Q_{h}-Q_{c}
\frac{dQ_{h}}{dt}=\frac{dQ_{c}}{dt} + \frac{dW}{dt}

The Attempt at a Solution

First of all I assumed
\frac{dW}{dt}=0
so
\frac{dQ_{h}}{dt}=\frac{dQ_{c}}{dt}= P_{h}

\int dQ_{h} = P_{h}\int dt = P_{h}t_{1} = ML_{f}
so
t_{1} = \frac{ML_{f}}{P_{h}}

but somehow to answer manages to squeeze out

t_{1} = \frac{ML_{f}T_{h}}{P_{h}T_{0}}

Whats the deal here? That answer doesn't seem to make much sense since it appears to be saying the hotter the heat reservoir, the longer it takes to melt the ice. I would have thought it to be the other way around.

And for part (b) I am not sure where to start.

 

[Mapes]

Where do you get dW/dt=0? Is the engine not producing power? It certainly should be, if it's perfectly efficient and there's a nonzero temperature difference between the two reservoirs.

 

[vladimir69]

I took it to mean W = constant, so d/dt W = 0
Still not sure how to proceed

 

[Mapes]

But W isn't constant, the total work you extract increases every second that the engine operates.

Think entropy: how much comes in from the hot reservoir? How much leaves with work? How much goes to the cold reservoir? How much is created? That's the key to analyzing Carnot engines.

 

[vladimir69]

I'm not really sure about the entropy of any of them
perhaps
dS_{h}=\frac{dQ_{h}}{T_{h}}
and maybe
dS_{c}=\frac{ML_{f}}{T_{c}}
No idea about how much entropy leaves as work.

 

[Mapes]

I'm not really sure about the entropy of any of them
perhaps
dS_{h}=\frac{dQ_{h}}{T_{h}}
and maybe
dS_{c}=\frac{ML_{f}}{T_{c}}

Great; although you're mixing differential and non-differential terms in the second equation, it's the right idea.

No idea about how much entropy leaves as work.

There is no entropy transfer associated with work.

 

[vladimir69]

I was just writing random equations not sure how this fits into the overall scheme of things