Thermal expansion of a rod with variable alpha

[Titan97]

Homework Statement

The coefficient of linear expansion of a rod of length 1 meter (at 300K) varies with temperature as $\alpha=\frac{1}{T}$, where T is the temperature. Find the increment in length when the rod is heated from 300K to 600K

Homework Equations

$$\Delta L=L\alpha \Delta T$$

The Attempt at a Solution

I considered a small element of length $dl$ at a distance of $l$ from one end.
Let the elements length increase by $dx$.
$$dx=dl\frac{2}{T}dT$$
How will I integrate the above expression

 

[haruspex]

dx=dl(2/T)dT

That does not balance in terms of infinitesimals. You can't equate one to a product of two.
There's no need to consider an element of length, the whole rod will expand in the same proportion.
Also, I didn't understand where the 2 came from. Should that be an alpha?

 

[Titan97]

I put the value of alpha. How can directly use $\Delta L =L\alpha z \Delta T$ when alpha varies with temperature? Shouldn't I integrate?

Also, if I take a length $l$, it will expand by $dl=dl\frac{2}{T}dT$.
Integrating, I got the answer as $2\ln 2$. But answer given is $3$.

 

[haruspex]

I put the value of alpha. How can directly use $\Delta L =L\alpha z \Delta T$ when alpha varies with temperature? Shouldn't I integrate?

Also, if I take a length $l$, it will expand by $dl=dl\frac{2}{T}dT$.
Integrating, I got the answer as $2\ln 2$. But answer given is $3$.

According to your original post, alpha is 1/T, but you seem to have substituted 2/T. From the given answer, 2/T is correct.
If you take L as the whole length, the increase in length, dL, equals LαdT, not dLαdT.
Please post your steps from there.

 

[Chestermiller]

Titan97: In terms of differential changes in length and temperature, the correct equation should read $dL=\alpha L dT$. In fact, this is the truly correct equation, and the equation $\Delta L=\alpha L \Delta T$ is only an approximation.

By the way, did you mean in your original post that $\alpha=1/T$, or did you mean $\alpha = 2/T$?

 

[Titan97]

OK. So $dL=L\alpha dT$
$$\int_L^{L+\delta L}\frac{dL}{L}=\int_{T_1}^{T_2}\frac{2}{T}dT$$
$$\ln\frac{L+\delta L}{L}=2\ln\frac{T_2}{T_1}$$

 

[Chestermiller]

OK. So $dL=L\alpha dT$
$$\int_L^{L+\delta L}\frac{dL}{L}=\int_{T_1}^{T_2}\frac{2}{T}dT$$
$$\ln\frac{L+\delta L}{L}=2\ln\frac{T_2}{T_1}$$

This is the same thing as $$\frac{L+\delta L}{L}=\left(\frac{T_2}{T_1}\right)^2$$
Can you see that?

 

[haruspex]

OK. So $dL=L\alpha dT$
$$\int_L^{L+\delta L}\frac{dL}{L}=\int_{T_1}^{T_2}\frac{2}{T}dT$$
$$\ln\frac{L+\delta L}{L}=2\ln\frac{T_2}{T_1}$$

Right. How to get rid of the logs?

 

[Titan97]

@haruspex I got the same expression as @Chestermiller 's post #7.
$$\ln(x)=2\ln(y)$$
$$\ln(x)=\ln(y^2)$$
$$x=y^2$$
$$1+\frac{\delta L}{L}=(\frac{T_2}{T_1})^2$$
$T_2/T_1=2$
$$1+\frac{\delta L}{L}=4$$
$$\frac{\delta L}{L}=3$$
Since $L=1\text{m}$
$$\delta L=3\text{m}$$

 

[haruspex]

Great. Please mark as solved!

 

[Titan97]

@haruspex I was going to mark it as solved today. But someone did that for me. I have also seen that some people can mark questions unsolved even if OP marked it as solved.

 

[Chestermiller]

@haruspex I was going to mark it as solved today. But someone did that for me. I have also seen that some people can mark questions unsolved even if OP marked it as solved.

Yes. I had marked it solved.