Calculating Max Work from Reversible Cooling of Liquid

In summary, the conversation discusses the calculation of maximum work that can be done on the surroundings while cooling a liquid with a constant molar heat capacity. The maximum work is limited by the second law of thermodynamics and cannot exceed the efficiency of a Carnot engine. The answer provided is 601 J.
  • #1
phizzle86
3
0

Homework Statement



One mole of a liquid with a constant molar heat capacity of 132 j/mol K is initially at a temp of 80 C. The heat capacity is independent of tempearture. Calculate the maximum work that could have been done onthe surroundings while cooling the liquid reversibly. The surroundings are at a temp of 25 C.


Homework Equations




Given: q= -7260 J , change in entropy= 22.3J/K for the cooling

Change entropy = (Cvdt)/T

The Attempt at a Solution



maximum work here is in a reversible process would be the total amount of heat that is released that has the potential to do work, so wouldnt' that b 7260 J? The answer provided is 601 J

thanks guys
 
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  • #2
Welcome to PF, phizzle86! :smile:

Did you consider that you cannot extract the maximum of energy as work?
The best you can do in any heat related process is the efficiency of a Carnot engine and in this case you can't even build one.
In other words, you need to respect the second law of thermodynamics.
 

1. What is reversible cooling of liquid?

Reversible cooling of liquid is a process in which a liquid is cooled down to a lower temperature without any significant increase in entropy or loss of energy. It is a thermodynamically reversible process that can be achieved by using a heat engine operating in a Carnot cycle.

2. How do you calculate the max work from reversible cooling of liquid?

The max work from reversible cooling of liquid can be calculated using the Carnot efficiency formula, which is equal to (T1-T2)/T1, where T1 is the initial temperature of the liquid and T2 is the final temperature after cooling. This efficiency can then be multiplied by the amount of heat extracted from the liquid to determine the max work.

3. What factors affect the max work from reversible cooling of liquid?

The max work from reversible cooling of liquid is affected by the temperature difference between the initial and final temperatures, the efficiency of the heat engine, and the amount of heat extracted from the liquid. Any losses in the system, such as friction or heat transfer to the surroundings, can also decrease the max work.

4. Can the max work be greater than the amount of heat extracted from the liquid?

No, the max work from reversible cooling of liquid cannot be greater than the amount of heat extracted from the liquid. This is due to the conservation of energy, which states that energy cannot be created or destroyed, only transferred or converted from one form to another.

5. How is calculating max work from reversible cooling of liquid useful in real-world applications?

Calculating the max work from reversible cooling of liquid is useful in understanding the efficiency of heat engines and in designing more efficient cooling systems. It can also be applied in various industries, such as refrigeration and air conditioning, where reversible cooling is used to maintain lower temperatures for preservation and comfort.

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