A copper cylinder is initially at $20.°C$. At what temperature will its volume be $0.150\%$ larger that it is at $20.0°C$?

Solution (first attempt):

I know that:


$\beta_{\text{copper}}=49.9\cdot 10^{-6}K^{-1}$

$V_{0}=k$ (here $k$ is any initial volume for the cylinder)

$\Delta V=1.00150\cdot V_{0}$

I don't know:


The formula to compute the increasing in volume is:

$$\Delta V=\beta V_{0}\Delta T$$


$$(1.00150\cdot k)=49.9\cdot 10^{-6}K^{-1}\cdot k\cdot(T_{f}-293.15K)$$

$$\frac{(1.00150\cdot k)}{49.9\cdot 10^{-6}K^{-1}\cdot k}=(T_{f}-293.15K)$$

$$\frac{(1.00150\cdot k)}{49.9\cdot 10^{-6}K^{-1}\cdot k}+293.15K=T_{f}$$

Does the way I have used to solve this problem is correct when a $k$ initial volume is given?

But... I when I do $k=100cm^3$ it gives me $20363.3K=20090.15°C$, which is a tremendous quantity.

  • $\begingroup$ Is $\beta$ the coefficient of linear expansion or the coefficient of volume expansion? $\endgroup$ – Chet Miller Feb 2 '17 at 18:01

You misinterpret the meaning of the fractional volume change when you write your formula. The change in volume is not 1.0015 Vo. This would mean a 100.15% increase in volume. You actually know that the relative change is 0.0015 so $$ \beta \Delta t =0.0015 $$ or $$ \Delta t =0.0015/\beta $$ which is about 30 degrees. There is no need to convert to Kelvin.


Congrats, you've successfully done a "sanity check," something that will make you a good student of physics in general! (I can't count the number of times that either I or a colleague has written a note in a margin of some student's exam like, "This mean-free-path for the electron is several times the size of the observable universe. Are you SURE that this sounds like a reasonable answer to you?")

Your problem in this case happens to be that you confused $\Delta V$ with $V_\text{final},$ so that $1.00150$ should probably be $0.00150,$ reducing the $\Delta T$ by three orders of magnitude. In other words you know that $\Delta V / V_0 = 0.00150,$ use that with the expression you have to find $0.00150/\beta = \Delta T,$ then add that temperature difference $\Delta T$ to $20^\circ\text{C}.$


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