How Much Pressure Does a Gallon of Freezing Water Exert

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Discussion Overview

The discussion revolves around the pressure exerted by a gallon of freezing water, particularly focusing on the relationship between temperature, pressure, and the phase transition from liquid to solid. The scope includes theoretical considerations and implications of pressure in confined systems.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the amount of pressure exerted by a gallon of freezing water.
  • Another participant explains that pressure along the liquid-solid coexistence line is independent of volume and depends on temperature, referencing the triple point of water and the phase diagram.
  • A participant elaborates on the pressure conditions required for freezing water in a confined vessel, suggesting that at near normal conditions, the pressure can be approximated as 10 MPa/K.
  • Further, they propose a scenario involving a cylinder of cold water with a piston, calculating the temperature required to freeze the water based on the weight applied to the piston.
  • Another participant shares links to external resources that may provide additional context on phase changes and forces involved.

Areas of Agreement / Disagreement

Participants express varying interpretations of the pressure-temperature relationship in the freezing process, with no consensus reached on the exact pressure exerted by a gallon of freezing water or the implications of the calculations presented.

Contextual Notes

The discussion includes assumptions about the conditions of the system, such as the nature of confinement and the specific weights applied, which may influence the freezing point and pressure calculations.

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How many pounds of pressure are exerted by a gallon of freezing water?
 
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How much would you like it to be? Pressure along the liquid-solid coexistence line is independent of volume; it depends only upon the temperature at which you establish the liquid-solid equilibrium. The liquid-solid coexistence begins at the triple point of water, 273.16 K and x Pa, proceeds to lower temperatures and higher pressures, 273.15 K and 0.1 MPa (ordinary freezing pt.), and goes wandering off through a fascinating phase diagram. At near normal conditions, you can figure something like 10 MPa/K for water in confined vessels --- once you cool to a point at which the pressure equals the burst or deformation limit for the vessel, the water freezes and the vessel bursts or deforms to accommodate the larger volume of the solid.
 
Originally posted by Bystander
How much would you like it to be? Pressure along the liquid-solid coexistence line is independent of volume; it depends only upon the temperature at which you establish the liquid-solid equilibrium. The liquid-solid coexistence begins at the triple point of water, 273.16 K and x Pa, proceeds to lower temperatures and higher pressures, 273.15 K and 0.1 MPa (ordinary freezing pt.), and goes wandering off through a fascinating phase diagram. At near normal conditions, you can figure something like 10 MPa/K for water in confined vessels --- once you cool to a point at which the pressure equals the burst or deformation limit for the vessel, the water freezes and the vessel bursts or deforms to accommodate the larger volume of the solid.

nice fact
just checking:
10 MPa/K is about 1000 tons per square meter per degree of temperature.
So I imagine I have a cylinder of cold water,
the bore of the cylinder is 1 square meter,
a 1000 ton weight rests on a piston in this cylinder
(so the pressure exerted on the water is about 10 MPa
above ambient)

If I wish to lift the weight by freezing the water in the cylinder
then according to your formula it will not freeze at 0 celsius
but has to be cooled to -1 celsius.

If I load the piston with 2000 tons then to get the water to
freeze I must chill it down to -2 celsius.
 
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