# Thermal expansion of a rod with variable alpha

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## 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
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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?

Gold Member
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
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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.

• Titan97
Chestermiller
Mentor
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
Gold Member
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
Mentor
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
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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?

Gold Member
@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}##
$$\delta L=3\text{m}$$

• Chestermiller
haruspex
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