Change in Entropy for Mixing of Two Liquids

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SUMMARY

The discussion focuses on calculating the change in entropy (ΔS) when mixing 1 liter of water at 20°C with 1 liter of water at 80°C. The relevant equation for this calculation is ΔS=∫(dQ/T) from state A to state B. Participants suggest assuming constant volume during the mixing process and using the specific heat capacity at constant volume (C_V) to derive the entropy change. The conversation highlights the complexity of the problem compared to ideal gas scenarios, emphasizing the need to ignore entropy contributions from the mixing process itself.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically entropy.
  • Familiarity with the concept of specific heat capacity (C_V).
  • Knowledge of integral calculus as applied to thermodynamic equations.
  • Basic principles of heat transfer in liquids.
NEXT STEPS
  • Study the derivation of the entropy change formula for mixing liquids.
  • Learn about the specific heat capacity of water and its implications in thermodynamics.
  • Explore the differences in entropy calculations between liquids and ideal gases.
  • Investigate the effects of temperature on the thermodynamic properties of water.
USEFUL FOR

Students in thermodynamics, physics enthusiasts, and anyone studying heat transfer and entropy in liquids will benefit from this discussion.

Alvin92SD
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Homework Statement


You mix 1 liter of water at 20°C with 1 liter of water at 80°C. Set up an equation to find the change in entropy for one of the volumes of liquid in terms of initial temperature (T1) and the temperature after the two volumes of water mixed (T2)


Homework Equations


ΔS=∫(dQ/T) from state A to state B



The Attempt at a Solution


Honestly I don't even know where to start or if I am using the right formula. I tried imagining a process where the volume of water is increased at constant temperature and then the temperature is increased from T1 to T2. I would be grateful if someone pointed me in the right direction.
 
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This is a bit tricky problem; it would be much more clear if you were dealing with ideal gases instead. Now you just need to ignore the increase of entropy coming from the mixing of the two liquids.

Perhaps you can assume that the volume of the liquid does not change when it's mixed and cooled/heated. Then you can write
dS = \left(\frac{\partial S}{\partial T}\right)_V dT +\left(\frac{\partial S}{\partial V}\right)_T dV = C_V \frac{dT}{T}
 

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