Three equal-length straight rods, of aluminum, Invar, and steel, all at $\mathbf{20}\mathbf{.0}\mathbf{}\mathbf{\text{C}}$, form an equilateral triangle with hinge pins at the vertices. At what temperature will the angle opposite the Invar rod be$\mathbf{59}\mathbf{.95}\mathbf{°}$? See Appendix E for needed trigonometric formulas and Table 18-2 for needed data.

When an object is heated or cooled, its length changes by an amount proportional to the original length and the temperature change. This process is called the linear expansion of the given substance. Thus, when the given rods arranged as a triangle are exposed to the heat, they expand linearly due to their coefficient of thermal expansion in one direction which is their increased length.

Formula:

The cosine law is given as:$c^2=a^2+b^2-2abcosC$.…(i)

The length expansion due to thermal radiation,$L=L_0(1+\alpha \Delta T)$ …(ii)

where$L_0$ is the original length of the body, is the coefficient of thermal linear expansion of the substance, and $\Delta T$is the temperature difference at both the ends of the body.

Applying cosine law from equation (i) to the equilateral triangle having $L_i,L_s\text{\hspace{0.17em}and}{L}_a,$we get

$L_i^2=L_a^2+L_s^2-2L_a{L}_{s}cos\theta$

Let${\alpha }_i,{\alpha }_s\text{and}{\alpha }_s$ be the coefficients of thermal expansion of the Invar, steel and aluminum sides$L_i,L_s\text{and}{L}_s$of the triangle.

Using equation (i), we get the above expression in the form of expansion as

$\begin{array}{c}[L_0{(1+{\alpha }_i\Delta T]}^2=[L_0{(1+{\alpha }_a\Delta T]}^2+[L_0{(1+{\alpha }_s\Delta T]}^2-2\left[L_0(1+{\alpha }_a\Delta T)L_0(1+{\alpha }_s\Delta T)cos\theta \right]\\ [{(1+{\alpha }_i\Delta T]}^2=[{\left(1+{\alpha }_a\Delta T)\right]}^2+[{\left(1+{\alpha }_s\Delta T)\right]}^2-2\left[(1+{\alpha }_a\Delta T)(1+{\alpha }_s\Delta T)cos\theta \right]\\ 1+{\alpha }_i^2\Delta T^2+2{\alpha }_i\Delta T\\ =(1+{\alpha }_a^2\Delta T^2+2{\alpha }_a\Delta T)+(1+{\alpha }_s^2\Delta T^2+2{\alpha }_s\Delta T)-2\left[(1+{\alpha }_a{\alpha }_s\Delta T^2+{\alpha }_a\Delta T+{\alpha }_s\Delta T)cos\theta \right]\end{array}$

Since${\alpha }_i^2\Delta T^2$, is negligible, we can ignore it; hence the equation becomes

$\begin{array}{c}1+2{\alpha }_i\Delta T=[(1+2{\alpha }_a\Delta T)+(1+2{\alpha }_s\Delta T)]-2[(1+{\alpha }_a\Delta T+{\alpha }_s\Delta T)cos\theta \\ 1+2{\alpha }_i\Delta T=[(2+2({\alpha }_a+{\alpha }_s)\Delta T)-2[(1+({\alpha }_a+{\alpha }_s)\Delta T)cos\theta \\ 1+2{\alpha }_i\Delta T=2[1+({\alpha }_a+{\alpha }_s)(1-cos\theta )\Delta T-cos\theta ]\\ \frac{1}{2}+{\alpha }_i\Delta T=[1+({\alpha }_a+{\alpha }_s)(1-cos\theta )\Delta T-cos\theta ]\\ ({\alpha }_a+{\alpha }_s)(1-cos\theta )\Delta T-{\alpha }_i\Delta T=cos\theta -1+\frac{1}{2}\\ \Delta T=\frac{(cos\theta -\frac{1}{2})}{[({\alpha }_a+{\alpha }_s)(1-cos\theta )-{\alpha }_i]}\\ =\frac{(\text{cos}(59.95°)-\frac{1}{2})}{\left[\begin{array}{l}(23\times {10}^{-6}{\text{/}}^0\text{C}+11\times {10}^{-6}{\text{/}}^0\text{C})(1-\cos (59.95°))\\ -0.7\times {10}^{-6}{\text{/}}^0\text{C}\end{array}\right]}\\ ={46.39}^0\text{C}\\ ~46°\text{C}\end{array}$

Then, the final temperature of the system is

$\begin{array}{c}{T}_f=T_i+\Delta T\\ ={20}^0\text{C}+{46}^0\text{C}\\ =66°\text{C}\end{array}$

Therefore, the temperature at which the angle opposite the Invar rodis of $59.95°$is $66°\text{C}.$