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 September 12th, 2015, 12:31 PM #1 Senior Member     Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty Another thermal expansion question Water at $\displaystyle 20^\circ C$ has volume $\displaystyle 55.50mL$. The water fills a container to the brim and when both are heated to $\displaystyle 60^\circ C$, $\displaystyle 0.35g$ of water is lost. What is the coefficient of volume expansion of the container?? $\displaystyle \bigtriangleup V = V_0\beta\bigtriangleup T$ $\displaystyle V_0 = 55.5\cdot10^{-3}L$ $\displaystyle \bigtriangleup T = 40^\circ C$ ($\displaystyle \bigtriangleup m\propto \bigtriangleup V$), $\displaystyle \bigtriangleup m = 0.35\cdot 10^{-3}kg = 0.35\cdot 10^{-3}L$ $\displaystyle \bigtriangleup V = V_0\beta\bigtriangleup T$ $\displaystyle \beta = \frac{\bigtriangleup V}{V_0\cdot T}$ $\displaystyle \beta = \frac{0.35\cdot 10^{-3}}{55.5\cdot 10^{-3}(40)}$ $\displaystyle \beta = 1.57\cdot 10^{-4}$ Book gives $\displaystyle \beta = 5.0\cdot 10^{-5}$ I think I've calculated $\displaystyle \beta$ for water and not the container?? But then the water and container initially have the same volume to start with. Both expand when heated and the water is lost through overflow and some surface vaporization? How to calculate $\displaystyle \beta$ for the container?? Thanks in advance. Last edited by hyperbola; September 12th, 2015 at 12:35 PM.
 September 14th, 2015, 05:02 AM #2 Senior Member     Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty Bump for a reply.
 September 15th, 2015, 02:59 AM #3 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,132 Thanks: 717 Math Focus: Physics, mathematical modelling, numerical and computational solutions Looks fine to me. I have no idea why the book gives a different result to yours.
 September 15th, 2015, 07:43 PM #4 Senior Member     Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty Okay. Then my next question is, if the calculation uses the change in temperature and change in volume of water, how is it that the calculated coefficient of expansion is the same for both the water and the container??? Ideally wouldn't we want a values for change in volume for the container in order to determine it's coefficient of expansion?? I just can't see how they would both have the same co-efficient
September 16th, 2015, 03:02 AM   #5
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 Originally Posted by hyperbola Okay. Then my next question is, if the calculation uses the change in temperature and change in volume of water, how is it that the calculated coefficient of expansion is the same for both the water and the container???
The amount of water lost is based on the thermal expansion of both. Your current solution assumes that the thermal expansion of the water is negligible.

Quote:
 Ideally wouldn't we want a values for change in volume for the container in order to determine it's coefficient of expansion?? I just can't see how they would both have the same co-efficient
Yes, or a value for the thermal expansion coefficient of water so you can calculate how much of the volume change was actually attributed to water. If the question gave you the thermal expansion of water, you could use that to calculate what the change in volume of the water is. Then you can compare that to the change in volume for the container + water to get the change in volume of the container.

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