Melting Ice with a Carnot Engine

[xinlan]

Homework Statement

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In 5 minutes of operation of the engine, the heat rejected by the engine melts a mass of ice equal to 3.50×10−2 kg .

Throughout this problem use Lf = 3.34*10^5 J/kg for the heat of fusion for water.

During this time, how much work W is performed by the engine?

Homework Equations

W = Q-ΔEint

The Attempt at a Solution

I calculated Q = M * Lf
3.50*10^-2 * 3.34*10^5 = 11690 J

then I don't know what to do..
please help me..

thanks in advance

 

[dynamicsolo]

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water.

The clue is in this sentence. The amounts of each form of water is large so that heat transfer does not cause their temperatures to change significantly. What are those temperatures? You will need them in order to calculate the efficiency e_c of this Carnot engine. Knowing that efficiency, you can now say that

e_c = W_net / Q_input .

What you have computed here

Q = M * Lf = 3.50*10^-2 * 3.34*10^5 = 11690 J

is the heat output (since it was used to melt some of the ice). You know from conservation of energy that

Q_input = W_net + Q_output .

So you should have everything you need to find W_net.

 

[ogb p]

I would first like to confirm that the information provided is accurate and relevant. Assuming that the Carnot engine is operating ideally and there are no other external factors affecting its performance, the calculation for the heat rejected by the engine seems correct.

To calculate the work performed by the engine, we can use the first law of thermodynamics: W = Q - ΔEint. Here, W represents the work done by the engine, Q represents the heat rejected by the engine, and ΔEint represents the change in internal energy of the engine.

Since the engine is operating in a cycle, the change in internal energy is zero. Therefore, the work done by the engine is simply equal to the heat rejected, which we have already calculated to be 11690 J.

In conclusion, the work performed by the Carnot engine in 5 minutes of operation is 11690 J. It is important to note that this is the maximum possible work that can be obtained from this engine, as it is operating at the highest possible efficiency (100%) for a Carnot engine.