A precise steel tape measure has been calibrated at 14°C. At 37°C, (a) will it read high or low, and (b) what will be the percentage error?

The initial temperature is \({T_ \circ } = {14^ \circ }{\rm{C}}\).

The final temperature is \(T = {37^ \circ }{\rm{C}}\).

The coefficient of linear expansion is \(\alpha = 12 \times {10^{ - 6}}\;{{\rm{/}}^{\rm{o}}}{\rm{C}}\).

The length of the tape increases when the temperature of the heat supplied to it increases.

For this problem, you have to use the formula\(L = {L_ \circ }\left( {1 + \alpha \Delta T} \right)\).

As the temperature increases, the length of the tape increases. Hence, the tape measure will be read low.

Let the tape measure a length \({L_ \circ }\) at temperature \({T_ \circ } = {14^ \circ }{\rm{C}}\) , and the reading of the same length be \(L\) at \(T = {37^ \circ }{\rm{C}}\).

The change in temperature is:

\(\begin{aligned}{c}\Delta T &= {37^ \circ }{\rm{C}} - 1{{\rm{4}}^ \circ }{\rm{C}}\\ &= {23^ \circ }{\rm{C}}\end{aligned}\)

Now, you get:

\(\begin{aligned}{c}L &= {L_ \circ }\left( {1 + \alpha \Delta T} \right)\\L - {L_ \circ } &= {L_ \circ }\alpha \Delta T\end{aligned}\)

Therefore, the percentage error is:

\(\begin{aligned}{c}\% {\rm{error}} &= \left( {\frac{{L - {L_ \circ }}}{{{L_ \circ }}}} \right) \times 100\% \\ &= \left( {\frac{{{L_ \circ }\alpha \Delta T}}{{{L_ \circ }}}} \right) \times 100\% \\ &= \alpha \Delta T \times 100\% \end{aligned}\)

Now, substituting the values in the above equation, you get:

\(\begin{aligned}{c}\% {\rm{error}} &= \alpha \Delta T \times 100\% \\ &= \left( {12 \times {{10}^{ - 6}}\;{{\rm{/}}^{\rm{o}}}{\rm{C}}} \right) \times \left( {{{23}^ \circ }{\rm{C}}} \right) \times 100\% \\ &= 2.8 \times {10^{ - 2}}\% \end{aligned}\)

Hence, the percentage error in the measurement is \(2.8 \times {10^{ - 2}}\% \).