Thermal Expansion of A Square Shaped Object

[Sunny007]

Suppose a square shaped object has an initial length of L1 and final length (after thermal expansion) of L2. Initial temperature is T1 and final temperature is T2. Suppose it has an area of A. So initial area is A1 and final area is A2 (after thermal expansion). Here A1 = (L1)^2 and A2 = (L2)^2 . Now area expansion coefficient B(beta) = (A2-A1)/A1*(T2-T1). or, B = (L2^2 - L1^2)/ L1*L1*(T2-T1) or, B = (L2-L1)(L2+L1)/ L1*L1*(T2-T1) or B = (L2-L1)*a/ L1 [ suppose alpha = a and we know that alpha = (L2-L1)/L1(T2-T1) ] or, 2a = (L2-L1)*a/ L1 or 2 = (L2-L1)/ L1 [ cancelling alpha from both sides) or, 2L1 = L2-L1 or L1= L2. So now does that mean that it has no expansion as Final length and initial length is equal?

Here is an image of the calculation.

[Mentor Note -- Image rotated to fix sideways orientation]

 

[Dale]

So now does that mean that it has no expansion as Final length and initial length is equal?

It means you made a mistake. In your 2nd equation (handwritten, please use LaTeX in the future) you have $2\alpha = \beta$ which is only true if $L_2=L_1$.

In general $$\beta =\left(1+\frac{L_2}{L_1} \right) \alpha$$ which again only reduces to $\beta=2\alpha$ in the special case $L_2=L_1$. So since you effectively assume that in your second equation it is no surprise that is what you get in the end when you simplify.

Edit: see below for a physical analysis by @kuruman. This post was purely an analysis of the posted algebra

 

[Sunny007]

It means you made a mistake. In your 2nd equation (handwritten, please use LaTeX in the future) you have $2\alpha = \beta$ which is only true if $L_2=L_1$.

In general $$\beta =\left(1+\frac{L_2}{L_1} \right) \alpha$$ which again only reduces to $\beta=2\alpha$ in the special case $L_2=L_1$. So since you effectively assume that in your second equation it is no surprise that is what you get in the end when you simplify.

Thank you so much sir for explaining this.

 

[kuruman]

You have to understand that the expression for linear thermal expansion is only an approximation because the change in length is proportional to the initial length and the ratio (or linear expansion coefficient) $\alpha = \Delta L/ L_0$ is of order 10^-5. Thus, anything will melt or catch fire before expanding to an appreciable fraction of its initial length.

The exact expression for the length $L$ of an object at temperature $T$ when its length is $L_0$ at $T_0$ is given by $$L(T)=L_0e^{\alpha (T-T_0)}.$$If that is expanded in power series, one gets$$L(T) =L_0\left[1+\alpha \Delta T+\frac{1}{2}(\alpha \Delta T)^2+\dots \right]$$We then note that each term in the expansion is about 10^-5 times smaller than the previous one, so we keep only the zeroth and the first order terms and write $$L(T) =L_0(1+\alpha \Delta T).$$ Now you can do either an exact calculation or an approximate calculation of the area but not mix the two.

Here is the exact calculation. We start with a $L_{01}\times L_{02}$ rectangle and write $$A(T)=L_{01}e^{\alpha \Delta T}\times L_{02}e^{\alpha \Delta T}=L_{01} L_{02}e^{\alpha \Delta T}e^{\alpha \Delta T}=A_0e^{2\alpha \Delta T}.$$If you want to write this as $A(T)=A_0e^{\beta \Delta T}$, then you need to define $\beta\equiv 2\alpha$.

For the approximate expression, you just expand the exact expression to first order as before to get $$A(T)\approx A_0 (1+\beta \Delta T)=A_0 (1+2\alpha \Delta T).$$If you mix exact and approximate expressions, as you seem to have done, then you need to be careful. If you call the expanded area the product of the two approximate sides, you have $$A(T)\approx L_{01}(1+\alpha \Delta T)\times L_{02}(1+\alpha \Delta T)=A_0\left[1+2\alpha \Delta T+(\alpha \Delta T)^2\right].$$ To first order, this reduces to the approximation obtained from the exact expression.

As you can see, you can always set $L_{01}=L_{02}$ anywhere in the exact or approximate calculations without any problem.

 

[Dale]

@Sunny007 note that @kuruman has actually analyzed the underlying physics for you, which I had seen decades ago but forgotten. My post just dealt with the algebra that you presented. I would recommend spending more time on his than on mine.