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I wanted to solve this problem:

In a glass there is 1 liter water at $5^\circ\ C$. How much water will spill out when the system is heated to $85^\circ\ C$?

Then I thought, OK I know $V_0$ which is 1 liter and $ΔT$ is $80^\circ\ C$, and I searched for the thermal coefficent in Google. It turns out it's varying. Look at this:

plot of thermal expansion coefficient vs temperature

So how am I going to calculate the thermal expansion? The coefficient varies from $1 × 10^{-4}$ to $6 × 10^{-4}$.

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You will have to set this problem up as an integral of dV/dT for both the water and the glass. Curve fit your graph(s), set up and solve the integral for both the glass and the water, determine the new volume of both, and do a subtraction to determine how much water spills out.

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  • $\begingroup$ Yes, but I haven't learned integral yet. But thank you to show off the solution :) $\endgroup$
    – agcgn
    Commented Dec 29, 2015 at 19:40
  • $\begingroup$ If you haven't learned how to integrate, your problem will be somewhat difficult. I could recommend a simple graphical method, but the concepts that it would take for the method to make sense come directly from integral calculus. $\endgroup$ Commented Dec 29, 2015 at 19:41
  • $\begingroup$ I see, you're right. Calculus is so important in physics. I just started highschool. I'll learn little by little then :) $\endgroup$
    – agcgn
    Commented Dec 29, 2015 at 19:53
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    $\begingroup$ @rabbit_hacker - I depends on the accuracy you need for the answer. If you want a rough figure of how much water will spill out, you can simply take the average of the thermal coefficient of expansion between 5C and 85C and use that for your calculation. $\endgroup$
    – user93237
    Commented Dec 29, 2015 at 20:46
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    $\begingroup$ @rabbit_hacker , I wish I had more students like you. I usually do not see your level of curiosity in the students that I am teaching. As you learn more math, and keep thinking about physics, you will begin to see the world from a somewhat different perspective than your peers, and I have no doubt that you will appreciate that perspective. Hang in there, and keep learning! $\endgroup$ Commented Dec 30, 2015 at 0:02

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