Two students each build a piece of scientific equipment that uses a 655 mm-long metal rod. One student uses a brass rod, the other an invar rod. If the temperature increases by $5.0°$C, how much more does the brass rod expand than the invar rod?

Two students each build a piece of scientific equipment that

uses a 655-mm-long metal rod.

The process of thermal expansion occurs when objects are heated as well as expanded. In case the temperature of an object changes through$∆T$, as well as length changes through $∆L$and the length's fractional changes $∆L/L$are effective to the temperature's change$∆T$.

The equation (18.8) in the form is

$\frac{∆L}{L}=\alpha ∆T\left(1\right)$

In this case, $\alpha$is the coefficient. Table 18.4 effectively gives the coefficient expression of linear for common materials, for instance

$\alpha =1.2\times {10}^{-5\circ }{C}^{-1}$

The equation for $∆L$is

$∆L=\alpha L∆T\left(2\right)$

In this case, L is the initial length. The temperature effectively changes through $∆T=5°C$for brass $∆{\alpha }_{brass}=1.9\times {10}^{-5°}{C}^{-1}$

For Brass, plug the values of ${\alpha }_{brass,}∆L,T$into equation (2) in order to get $∆L_{brass}$

$\begin{array}{rcl}∆L_{brass}& =& {\alpha }_{brass}L∆T\\ & =& (1.9\times {10}^{-5°}{C}^{-1})(655mm)\left(5^{°}C\right)\\ & =& 0.06222mm\end{array}$

plug the values of $∆L,T,{\alpha }_{invar}$ into equation (2) in order to get $∆L_{invar}$

role="math" localid="1649070713758" $\begin{array}{rcl}∆L_{invar}& =& {\alpha }_{invar}L∆T\\ & =& (0.09\times {10}^{-5°}{C}^{-1})(655mm)\left(5^{°}C\right)\\ & =& 0.00295mm\end{array}$

The calculation depicted the brass changes through a larger value of invar and the difference is

role="math" localid="1649070956077" $\begin{array}{rcl}∆L& =& ∆L_{brass}-∆L_{invar}\\ & =& 0.06222mm-0.00295mm\\ & =& 0.05927mm\\ & =& 5.927\times {10}^{-2}mm\end{array}$

Therefore, the answer is$\begin{array}{rcl}& & 5.927\times {10}^{-2}mm\end{array}$