# Thermodynamics in mercury filled tube

Please provide a solution for this problem.

@Damir @Miras помните задачу 1.6 термодинамики Овчинкина?)) Попробуйте решить эту тогда)

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So this problem is taken from Овчинкина?

Хорошая задача, чтобы вспомнить
Uh, I remember the second part of this problem.
Firsly, we should find a central mass of metal tube.

L=l\left(1- \frac{x}{2} \right)

We know that wen we increase the temperature of metal tube, it’s own lengh and mercury volume are increasing, but we need chose a x, when it does not depend to temperature. It takes this form:

\frac{dL}{dl}=0

Be careful, because we need only find how length of mercury changes depending on length of metal tube. In other words \displaystyle\frac{dx}{dl} We don’t need to find single expression for temperature, because l replaces the effect of temperature (If someone understands me). This is like a function .

\frac{d(1 \cdot l)}{dl}-\frac{1}{2}\frac{d(l \cdot x)}{dl}=1-\frac{x}{2}-l\frac{dx}{dl}=0

Phew, I hope that, I wrote correctly.
So, we considered all phisics there))
Now, let’s an increase of mercury’s volume and metal’s length. Since the increase in the pipe is equilateral, the increase is three times more. So:

l_1=l(1+\beta T)\\ V_1=V(1+3\beta T)\\ v_1=v(1+ \alpha T)\\

We know that x is the part of metal tube with mersure and we can find a connertion between their volumes. It is enough to find x.

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What value of x are you getting?

x=0.125
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Ok you are gettin’ the correct answer. The answers given are:

1. 1/8
2. 0.131 or 0.963
3. Some long algebraic expression

It has three possible answers?

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No, the answer of first part is 0.125, the answer of second part is 1. 0.131 or 0.963 and the answer of third part is a large algebraic equation.

You have found out the answer of first part as 0.125 which is indeed correct. Please solve the other 2 parts as well.

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Oh, understandble, have a nice day.
I forgot the solution of 2 part, I think we must find errors of x.

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