Do I need to use simultaneous equations for this problem (below), and if so how do I derive them? If not, how do I find out the unknown variable; steel rulers length using the length given for the rod. Thanks.;

At 25°C, a rod is exactly 23.89 cm long on a steel ruler. Both the rod and the ruler are placed in an oven at 220°C, where the rod now measures 24.01 cm on the same ruler. What is the coefficient of thermal expansion for the material of which the rod is made? The linear expansion coefficient of steel is 11 x 10^{-6} 1/C°.

Stu
Oct 1, 2014

#5**+11 **

Hi Alan,

I don't know about this stuff so I am just using your formula and a bit of logic (most likely incorrect)

"If α is the linear coefficient of thermal expansion then

α = change in length/(original length*temperature change)"

$$\\11\times 10^{-6}=\frac{\triangle l_s}{23.89\times 195}\\\\

\triangle l_s = 11\times 10^{-6}\times23.89\times 195 \\\\$$

$${\mathtt{11}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{6}}\right)}{\mathtt{\,\times\,}}{\mathtt{23.89}}{\mathtt{\,\times\,}}{\mathtt{195}} = {\mathtt{0.051\: \!244\: \!05}}$$

$$\\\alpha_R=\frac{\triangle l_s+(24.01-23.89)}{23.89*195}\\\\

\alpha_R=\frac{0.05124405+0.12}{4658.55}\\\\$$

$${\frac{\left({\mathtt{0.051\: \!244\: \!05}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.12}}\right)}{{\mathtt{4\,658.55}}}} = {\mathtt{0.000\: \!036\: \!759\: \!088\: \!128\: \!3}}$$

$$\\\alpha_R=3.7\times 10^{-5} $ per degree celcius$ \qquad \\

$(correct to 2 significant figures)$$$

Is this correct?

Melody
Oct 1, 2014

#2**+8 **

Ok, so the ruler can be any length so long as the length chosen ends up bigger than the ending rod length. I think i have it.

Stu
Oct 1, 2014

#3**+8 **

I set the ruler to 50cm, and found that the ruler grew 0.10725 (please check), then I found the ratio which was (ruler final by ruler initial) 1.002145 (please check I'm sure this is way to big of a number to be acurate), then I got stuck to calculate the rods actual growth and solve for the materials co-eficient of thermal expansion. Please assist me.

Stu
Oct 1, 2014

#4**+8 **

If α is the linear coefficient of thermal expansion then

α = change in length/(original length*temperature change)

Note: Edited and corrected thanks to Melody's more carefully thought through analysis below!

Alan
Oct 1, 2014

#5**+11 **

Best Answer

Hi Alan,

I don't know about this stuff so I am just using your formula and a bit of logic (most likely incorrect)

"If α is the linear coefficient of thermal expansion then

α = change in length/(original length*temperature change)"

$$\\11\times 10^{-6}=\frac{\triangle l_s}{23.89\times 195}\\\\

\triangle l_s = 11\times 10^{-6}\times23.89\times 195 \\\\$$

$${\mathtt{11}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{6}}\right)}{\mathtt{\,\times\,}}{\mathtt{23.89}}{\mathtt{\,\times\,}}{\mathtt{195}} = {\mathtt{0.051\: \!244\: \!05}}$$

$$\\\alpha_R=\frac{\triangle l_s+(24.01-23.89)}{23.89*195}\\\\

\alpha_R=\frac{0.05124405+0.12}{4658.55}\\\\$$

$${\frac{\left({\mathtt{0.051\: \!244\: \!05}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.12}}\right)}{{\mathtt{4\,658.55}}}} = {\mathtt{0.000\: \!036\: \!759\: \!088\: \!128\: \!3}}$$

$$\\\alpha_R=3.7\times 10^{-5} $ per degree celcius$ \qquad \\

$(correct to 2 significant figures)$$$

Is this correct?

Melody
Oct 1, 2014