Chaqpter 16 - Temperature and Heat - Problems and Conceptual Exercises - Page 567: 22

$0.0014\ in$

We know that $\Delta d_{steel}-\Delta d_{copper}=(\alpha_{steel}-\alpha_{copper})d_{\circ}\Delta T$ We plug in the known values to obtain: $\Delta d_{steel}-\Delta d_{copper}=(1.73\times 10^{-5}-1.70\times 10^{-5})(8.0)(610-22)=0.0014\ in$