If I have a sealed enclosure full of water (constant volume) at 25˚C at atmospheric pressure, I then heat the water to 50˚C. Would the pressure in the sealed enclosure change?
If the pressure has changed, how would I go about calculating the change?
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Yes, at constant density, the pressure increases as the temperature does:
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For example, having water sealed at atmospheric pressure at $4\sideset{^{\circ}}{}{\mathrm{C}}$ will have a density of approximately $1 \frac{\mathrm{g}}{\mathrm{cm}^3}$. If we increase the temperature to $30\sideset{^{\circ}}{}{\mathrm{C}}$, maintaining the density (since the enclosure is sealed), the pressure will rise up to $100 \, \mathrm{bar}$.
Find equations describing the rate of change here.
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2Guille's answer is the correct answer if the original poster meant literally constant volume (i.e. the water is enclosed in a hypothetical container that is infinitely rigid). Most real containers, however, would easily strain slightly to accommodate the < 1% increase in water volume without inducing much stress, hence resulting in negligible pressure change. – Sean49 Oct 17 '17 at 06:37
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Completely agree with you. This would only happen with an hypothetical container. To calculate the rate of change we would have to describe the geometry of the container (preferible a sphere) and its Stiffness. To make a fine estimation, the equations to be solved can be quite complex. – Guille Oct 17 '17 at 06:53
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I completely disagree. A 1% compression of liquid water would cause an enormous pressure. At room temperature, the bulk modulus of water is 2150 MPa, so a 1% compression would cause a pressure of 21.5 MPa. – Chet Miller Oct 17 '17 at 13:17
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2If you filled a plastic water bottle with 25 C water to the very brim and sealed it with a pressure gauge inside, then heated the water to 50 C or even 99 C, that pressure gauge would not read 21.5 MPa higher. My point was specifically because it requires enormous pressure to compress liquid by 1%, in practice the water would not be compressed 1% and would instead strain the container so its inner volume expanded by 1%. For most materials and geometries of most containers, the amount of pressure required to strain its volume to 101% is tiny. This would determine the internal pressure – Sean49 Oct 17 '17 at 18:54
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If the volume is constant (the container is very stiff/rigid), the rise in pressure would be dramatic indeed. In fact, that is the whole purpose of involving pressure tanks in all enclosed fluid circuits; otherwise the rise in pressure would quickly trigger the safety valve (hopefully present), and if absent, damage the installation.
P Marecki
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The OP is asking about a rigid, liquid full container, not a container with air pressure on top of a diaphragm. – David White Aug 29 '20 at 01:17
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Yes, I know; that's why @Guille's answer is correct here. I wanted to merely point out here, that the effect the OP is asking about is extremely common, and is the reason for the need of inlcuding the pressure tanks into every closed/rigid circulation system. – P Marecki Aug 29 '20 at 07:55
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Yes the pressure would increase, but the amount would depend on the material of the container. Example, a water balloon vs a steel drum expand differently. It should also depend if there is any gas present, or only 100% water. Could also depend if container expansion is only elastic and not plastic deformation as well.
Scott
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To a very good approximation, liquid water can be treated as incompressible, and therefore the pressure would not have increased. In practice, the pressure would be determined by the ambient pressure outside of the container, because no container is infinitely rigid.
An exception to this would be if the ambient pressure were low enough such that $50\sideset{^{\circ}}{}{\mathrm{C}}$ were above the boiling point of water at that pressure. In that case the water would try to boil. If the container were rigid enough to contain the water, the pressure would increase until the boiling point at that new pressure were $> 50\sideset{^{\circ}}{}{\mathrm{C}}$. If the container were not rigid enough (to support the pressure difference between that internal pressure and whatever the outside ambient pressure is), then it would rupture and fail.
