September 12th, 2015, 11:31 AM  #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 11:35 AM. 
September 14th, 2015, 04:02 AM  #2 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty 
Bump for a reply.

September 15th, 2015, 01:59 AM  #3 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,111 Thanks: 706 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, 06:43 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 coefficient 
September 16th, 2015, 02:02 AM  #5  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,111 Thanks: 706 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
Quote:
 

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expansion, question, thermal 
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