Calculating Maximum Work from Cooling Liquid Reversibly

I don't think this summary is correct. The conversation is discussing how to calculate the maximum work that can be done on the surroundings while cooling a liquid with a constant molar heat capacity of 132 j/mol K from 80 C to 25 C. The equation given is q=-7260 J and the change in entropy is 22.3 J/K. The person is having trouble understanding what maximum work means and is questioning if it is 7260 J. The answer provided is actually 601 J, which is calculated using the first law of thermodynamics and the fact that Cp is constant.
  • #1
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



I guess my problem here is not understanding what maximum work is, i would think that the 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
 
Physics news on Phys.org
  • #2
I'm not sure this is helpful because I don't see how the answer of 601 J was arrived at, but from first law ΔU=Q-W where ΔU=n∫Cv*dt. For a liquid Cv≈Cp but Cp is constant therefore ΔU=n*Cp*ΔT. Maximum work would occur if Q=0 (insulated container). Therefore Wmax=ΔU

Maybe I am missing something
 

What is the definition of maximum work in cooling liquid?

Maximum work in cooling liquid refers to the maximum amount of usable energy that can be extracted from a cooling liquid as it is cooled reversibly, without any losses or inefficiencies.

How is maximum work calculated from cooling liquid?

The formula for calculating maximum work from cooling liquid is Wmax = ΔH - TΔS, where ΔH is the enthalpy change of the liquid, T is the temperature at which the liquid is cooled, and ΔS is the entropy change of the liquid.

What factors affect the maximum work in cooling liquid?

The maximum work in cooling liquid is affected by the temperature at which the liquid is cooled, the properties of the liquid, and the efficiency of the cooling process.

What is the significance of calculating maximum work from cooling liquid?

Calculating maximum work from cooling liquid allows scientists and engineers to determine the maximum amount of energy that can be harnessed from a cooling process, which can help optimize the design and efficiency of cooling systems.

Can maximum work be achieved in real-world cooling processes?

No, it is not possible to achieve maximum work in real-world cooling processes. This is because there will always be some level of inefficiency or energy loss in the cooling process, making it impossible to extract the maximum amount of work from the cooling liquid.

Suggested for: Calculating Maximum Work from Cooling Liquid Reversibly

Replies
8
Views
778
Replies
6
Views
564
Replies
4
Views
1K
Replies
9
Views
787
Replies
2
Views
49
Replies
3
Views
642
Back
Top