I Resulting temperature of mixed liquids

AI Thread Summary
The discussion centers on understanding the energy dynamics when mixing two liquids at different temperatures and how this relates to thermodynamics and entropy. It highlights that simply averaging the temperatures of the two liquids does not account for the energy available for a heat engine, as the final temperature will differ when an engine is involved. The conversation emphasizes the importance of specific heat capacity in calculating the energies of the liquids and determining the resulting temperature after mixing. Additionally, it acknowledges the role of entropy in these processes, indicating that the energy from the temperature difference is retained in the mixed state. Overall, the participants are revisiting foundational concepts in thermodynamics to clarify these relationships.
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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