Liquids, internal energy and specific heat capacity

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Discussion Overview

The discussion revolves around the relationship between internal energy, work done, and heat capacity in the context of an adiabatic process involving a liquid. Participants explore the implications of the first law of thermodynamics and the definitions of heat and internal energy in this scenario.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant states that in an adiabatic process, the heat transferred (dQ) is zero, leading to the conclusion that the work done on the system is equal to the change in internal energy (dU).
  • Another participant argues that the correct relationship for heat transfer is Q = mass times specific heat times temperature rise, emphasizing that work done is converted to heat through interparticulate friction.
  • Some participants propose that for an incompressible liquid, the relationship dU = m*s*dT holds true, suggesting that this can be used to relate internal energy changes to temperature changes.
  • There is a challenge regarding the standard textbook definitions of heat capacities, with participants questioning the correctness of defining heat capacities in terms of internal energy (dU) instead of heat (Q).
  • One participant suggests that defining heat capacities in terms of Q is a simplification for beginners and references resources for further reading on thermodynamics.

Areas of Agreement / Disagreement

Participants express differing views on the definitions and relationships between heat, work, and internal energy. There is no consensus on whether heat capacities should be defined in terms of dU or Q, and the discussion remains unresolved regarding the correct formulation.

Contextual Notes

Participants note that the definitions and relationships discussed may depend on the specific conditions of the system, such as whether the liquid is incompressible or undergoing deformation. There are also references to potential limitations in the standard definitions found in textbooks.

metalrose
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A liquid contained in an adiabatic container is shaked vigorously so that it its temp. Increases.

The heat capacity for the liquid is given, the rise in temp. Is given.

According to the first law of thermo, dQ=dW + dU
here dQ is 0.

Asked, is to find the work done on the system, i.e. -dW =dU

The correct solution to this is dU = m*s*dT

My question is , how can this formula be used ?? Since the correct formula is dQ=m*s*dT
??
 
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No the correct formula is not q = msΔT.

q is the heat transferred into the system and this is defined to be zero by the adiabatic nature of the process.

Work is transferred into the system.
Within the system this work is converted to heat by interparticulate friction.
This heat increases the temperature of the system.

The correct equation connecting tmeperature rise to heat used to effect that temperature rise is

Q = mass times specific heat times temperature rise.

Q is numerically (and factually) equal to the work transferred into the system.

You should remember that q and w are always the energy that crosses the boundary into or out of the system when used in the first law.
The internal energy is a sort of book keeping account to balance the inputs and outputs.
 
metalrose said:
A liquid contained in an adiabatic container is shaked vigorously so that it its temp. Increases.

The heat capacity for the liquid is given, the rise in temp. Is given.

According to the first law of thermo, dQ=dW + dU
here dQ is 0.

Asked, is to find the work done on the system, i.e. -dW =dU

The correct solution to this is dU = m*s*dT

My question is , how can this formula be used ?? Since the correct formula is dQ=m*s*dT
??

Actually, the more basic (and correct) formula for an incompressible liquid is dU = m*s*dT. If you add heat to an incompressible liquid that is not being deformed, you get dU = m*s*dT = dQ, and this gives you a relationship between the amount of heat added to the temperature change. On the other hand, if you do work to deform an incompressible liquid without adding heat, you get dQ = 0, but still have dU = m*s*dT.
 
Chestermiller said:
Actually, the more basic (and correct) formula for an incompressible liquid is dU = m*s*dT. If you add heat to an incompressible liquid that is not being deformed, you get dU = m*s*dT = dQ, and this gives you a relationship between the amount of heat added to the temperature change. On the other hand, if you do work to deform an incompressible liquid without adding heat, you get dQ = 0, but still have dU = m*s*dT.

I think that would be correct. I was thinking along the same lines initially. But the standard textbook definition of heat capacities involve heat, i.e. Q and not U. Are you sure your formulation is correct?

And if it is, could you point me to some textbooks or resources where I could find heat capacities defined in terms of dU ?
 
metalrose said:
I think that would be correct. I was thinking along the same lines initially. But the standard textbook definition of heat capacities involve heat, i.e. Q and not U. Are you sure your formulation is correct?

And if it is, could you point me to some textbooks or resources where I could find heat capacities defined in terms of dU ?

Yes. Defining it in terms of Q is just for beginners.

See Amazon under the key word Thermodynamics

Smith and Van Ness

Van Ness and Abbot
 

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