Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: R717NOVAH on September 12, 2020, 10:40:28 PM
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I can witness liquid NH3 achieving sub 'boiling point' temperatures by simpling removing a thermocouple from a container of liquid, typically around -40 to -60°C.
I see that as the Ammonia evaporates energy is lost from the liquid droplets and I would imagine a loss in density.
With the exception of mathematical explanations from the works of Geoffrey D. Kaiser & Richard F. Griffiths, I am yet to find an explanation that hits home of how this occurs other than terms like entrained air and adiabatic volumetric expansion, partial pressures etc. I would like to find some visualisation of what is exactly going on in this process.
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Look at the ammonia phase diagram.
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Look at the ammonia phase diagram.
The P/T of Ammonia at equilibrium indicates - 33°C am I missing something, how does it drop below this point without a drop in pressure?
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You have to cool the ammonia - otherwise, the temperature will not drop. At a pressure of 1 Atm. liquid ammonia exists in the temperature range of ~-75 to -33°C.
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You have to cool the ammonia - otherwise, the temperature will not drop. At a pressure of 1 Atm. liquid ammonia exists in the temperature range of ~-75 to -33°C.
This is the object of my question - how is the Ammonia being cooled below -33°C ?
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Liquid nitrogen will even solidify ammonia.
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You have to cool the ammonia - otherwise, the temperature will not drop. At a pressure of 1 Atm. liquid ammonia exists in the temperature range of ~-75 to -33°C.
This is the object of my question - how is the Ammonia being cooled below -33°C ?
In this case there is no cooling taking place other than the evaporative cooling, the liquid ammonia, sitting in an adiabatic container can be measured at -33, when removing the thermometer it will drop significantly. How do I calculate this?
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when removing the thermometer it will drop significantly
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I doubt it.
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I can witness liquid NH3 achieving sub 'boiling point' temperatures by simpling removing a thermocouple from a container of liquid, typically around -40 to -60°C.
Can you describe the setup/process/observations more clearly? Are we talking about a real experiment/setup?
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Evaporation continues below the boiling point, only without bubbles, as we often observe with water. So, further cooling could occur.
Though, from -33°C to -70°C is doubtful, and "by removing the thermometer" sounds like a observation gone wrongly.
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In this case there is no cooling taking place other than the evaporative cooling, the liquid ammonia, sitting in an adiabatic container can be measured at -33, when removing the thermometer it will drop significantly. How do I calculate this?
You mean the thermometer temperature drops, not the ammonia in the container? (How would you measure that?) Sounds like drops of liquid ammonia on the thermocouple evaporate, taking their latent heat of evaporation from the thermocouple. (Same reason you cool down when you sweat.) If the drops are in contact with a metal thermocouple, they will draw the heat from there rather than the atmosphere. This isn't an equilibrium situation.
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Evaporation continues below the boiling point, only without bubbles, as we often observe with water. So, further cooling could occur.
Though, from -33°C to -70°C is doubtful, and "by removing the thermometer" sounds like a observation gone wrongly.
Thank you for your response. So we are simply talking about evaporative cooling nothing complex and I am over thinking the entire situation. _ I am intreged about what you said "only without bubbles" what does this mean?
I am a Refrigeration engineer, when we observe Ammonia outside of equilibrium simply in a beaker, we can measure -33°C which satisfys P/T relationship, once we remove the thermometer the Ammonia evaporates and cools down to -60 rapidly sometimes momentarily -70°C or less. If the beaker or container is not insulated the entire container will usually drop to around -45°C Note this is not controlled scientific observations just anecdotal.
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Without bubbles: just like water makes bubbles above +100°C, but still evaporates below +100°C, without bubbles.
But -70°C is damn cold. Whether evaporation achieves this, and quickly?