Solving Volumetric Expansion of Water in Glass

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The discussion centers on calculating the volumetric expansion of water and glass when the temperature is lowered from 100°C to 20°C. The coefficients of volume expansion are 2.7 x 10-5 K-1 for glass and 2.1 x 10-4 K-1 for water. The calculations show that the glass shrinks by 0.00216 m3 while the water shrinks by 0.0168 m3, resulting in a net increase of 0.01464 m3 of water that can be added. The discussion also emphasizes the importance of using appropriate volume assumptions in physics problems.

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Homework Statement


An ordinary glass is filled to the brim with water at 100oc. How much water could be added to the glass if the temperature is lowered to 20oC? Assume that the coefficient of volume expansion for glass is 2.7 x 10-5 K-1 and for water it is 2.1 x 10-4 K-1.

Homework Equations



ΔV = γ⋅VΔθ

The Attempt at a Solution



Δθ = 80 K
I will assume 1m3 for volume to make the calculations simple for both.

ΔVglass = 2.7 x 10-5 K-1⋅1m3⋅80k = 0.00216m3

ΔVwater = 2.1 x 10-4 K-1 ⋅ 1m3⋅80k = 0.0168m3

I reasoned that since there is differential shrinkage, the glass shrinks 0.00216m3, thus there is less available room for storing water, but the water shrank also, if we subtracted the results from each other, we would have the net change. So we could add 0.01464m3 more water.

I wanted to check my supposition that I could use 1m3 for both the glass and the water. The glass does not occupy a full cubic meter, since its mostly walls with a void in side. Is that right? Or does that not get addressed in intro physics? Have I worked out the problem correctly or ? I probably should report this in percentage as 1.46% change, no?

Thanks!
 
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chopnhack said:

Homework Statement


An ordinary glass is filled to the brim with water at 100oc. How much water could be added to the glass if the temperature is lowered to 20oC? Assume that the coefficient of volume expansion for glass is 2.7 x 10-5 K-1 and for water it is 2.1 x 10-4 K-1.

Homework Equations



ΔV = γ⋅VΔθ

The Attempt at a Solution



Δθ = 80 K
I will assume 1m3 for volume to make the calculations simple for both.

ΔVglass = 2.7 x 10-5 K-1⋅1m3⋅80k = 0.00216m3

ΔVwater = 2.1 x 10-4 K-1 ⋅ 1m3⋅80k = 0.0168m3

I reasoned that since there is differential shrinkage, the glass shrinks 0.00216m3, thus there is less available room for storing water, but the water shrank also, if we subtracted the results from each other, we would have the net change. So we could add 0.01464m3 more water.

I wanted to check my supposition that I could use 1m3 for both the glass and the water. The glass does not occupy a full cubic meter, since its mostly walls with a void in side. Is that right? Or does that not get addressed in intro physics? Have I worked out the problem correctly or ? I probably should report this in percentage as 1.46% change, no?

Thanks!
Hint: Suppose you replace the water in the full glass with an equivalent volume of glass. Won't the glass remain full, no matter the temperature ?
 
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If this was all glass, then the entire piece would expand and contract together.

What were you hinting at? Did I get it wrong?
 
chopnhack said:
If this was all glass, then the entire piece would expand and contract together.

What were you hinting at? Did I get it wrong?
You seemed rather uncertain as to whether your supposition was correct so I tried to give a hint to help your confidence.
chopnhack said:
I wanted to check my supposition that I could use 1m3 for both the glass and the water. The glass does not occupy a full cubic meter, since its mostly walls with a void in side. Is that right? Or does that not get addressed in intro physics? Have I worked out the problem correctly or ? I probably should report this in percentage as 1.46% change, no?
You could have used something more general for the volume, such as V0, or V100. The main point, however, is to use this for both glass and water.

By the way:
In the blue bar above the window for composing a post, find X2 and X2 for superscript and subscript. Also find there ∑ to give a large set of symbols, including the degree ° symbol.

Even better is to learn a little LaTeX .

Added in Edit: Your result looks correct to me.
 
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Gotcha, thank you for confirming the work. I hope you can understand my trepidation - the glass itself doesn't occupy a cubic meter, so when I made that assumption, I was concerned that I had missed something fundamental. I had copied and pasted the question and didn't go back to revise it. I am familiar with the site tools now, but latex still gives me a rash ;-)
cheers
 
chopnhack said:
I probably should report this in percentage as 1.46% change
As I read the question, it is asking for an actual volume. (If you quote a percentage, would that be of the original volume or of the new volume?)
 
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haruspex said:
As I read the question, it is asking for an actual volume. (If you quote a percentage, would that be of the original volume or of the new volume?)
I wrote it up as the ability to add ~1.4% more water. I would take that as an increase over the original volume.
 

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