Resulting temperature of mixed liquids

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SUMMARY

The discussion focuses on the thermodynamic principles governing the mixing of two liquids at different temperatures and the implications for energy and entropy. Specifically, it highlights that the resulting temperature of the mixture is not simply the average of the two initial temperatures due to the presence of energy E, which is derived from the temperature difference. The specific heat capacity of water, approximately 4.18 J/g, is mentioned as a critical factor in calculating the energies of the liquids and determining the final temperature of the mixture. The conversation emphasizes the importance of understanding entropy in this context, particularly when a heat engine is involved.

PREREQUISITES
  • Understanding of thermodynamics principles
  • Knowledge of specific heat capacity, particularly for water (4.18 J/g)
  • Familiarity with energy calculations in thermal systems
  • Basic concepts of entropy and its effects on thermal processes
NEXT STEPS
  • Research the equation for change in internal energy as a function of heat capacity and temperature change
  • Study the relationship between entropy and temperature in thermodynamic systems
  • Learn about the principles of heat engines and their efficiency
  • Explore advanced thermodynamic concepts such as enthalpy and its applications
USEFUL FOR

Students and professionals in physics, engineering, and thermodynamics, particularly those interested in heat transfer, energy calculations, and the principles of entropy in thermal systems.

synch
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[ I think I should have asked this many years ago when I was studying !!! However it will help me re-understand thermodynamics and entropy a bit, and may be good for others on the same path so... ]

Considering two portions of a liquid. Container A liquid is at a high temperature, container B is at a low temperature. The temperature difference could be used to drive a heat engine, so there is an calculable amount of energy E available that results from the difference in temperatures and heat contents.

Mixing the two (same volumes) liquids results in the temperature changing to a point somewhere between the starting values. If I simply assume the result is just the average, what has happened to the energy E ? So, I guess that assumption is wrong and the energy E is still present, the temperature will be higher than the simple average.

How is energy E then accounted for, and the temperature calculated ? I guess that is the whole gist of entropy !?
 
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Hint: how do you calculate the energy of each fluid as a function of temperature?
 
... { OMG I am starting to remember uni from forty years ago... ] that would be using the specific heat capacity of the liquid ,,,errrm say 4.18 J / g water as an approximate constant...so calculate the energies of the two portions, add them and then use the combined amount to work back to a temperature of the mix...
will that give the correct temperature ? So, the energy that would have been available for a heat engine, is simply contained in the mixed whole I guess ? Is there any entropy effect ?
 
synch said:
... { OMG I am starting to remember uni from forty years ago... ] that would be using the specific heat capacity of the liquid ,,,errrm say 4.18 J / g water as an approximate constant...so calculate the energies of the two portions, add them and then use the combined amount to work back to a temperature of the mix...
will that give the correct temperature ? So, the energy that would have been available for a heat engine, is simply contained in the mixed whole I guess ? Is there any entropy effect ?
Yes, there is an entropy effect. In the case without the engine, the final temperature is the average temperature. In the case where there is an engine involved, the final temperature is not the average of the initial temperatures.

Do you remember the equation for the change in the internal energy as a function of the heat capacity and the temperature change?
 
Chester knows more than me, so I’ll let him handle this.
 
It was something involving enthalpy and T delta S.... but before I waste your time, thanks guys , I'll read up a bit and get back
 

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