(II) To what temperature would you have to heat a brass rod for it to be 1.5% longer than it is at 25°C?

• The coefficient of thermal expansion of brass is\({\alpha _{br}} = 19 \times {10^{ - 6}}\;{\rm{/^\circ C}}\).
• The ratio of the change in the length of the brass rod to the original length is\(\frac{{\Delta {l_{br}}}}{{{l_{br}}}} = 1.5\% = 1.5\left( {\frac{1}{{100}}} \right) = 0.015\).
• The initial temperature of the brass rod \({T_i} = 25^\circ {\rm{C}}\).

The change in the temperature of the brass rod is the ratio of the change in its length to the product of the thermal expansion coefficient and the initial length of the rod. The final temperature of the brass rod is dependent on its initial temperature, change in length, thermal coefficient, and initial length.

The change in the length of the brass rod can be expressed as

\(\begin{aligned}{c}\Delta {l_{br}} &= {\alpha _{br}}{l_{br}}\Delta {T_{br}}\\\Delta {l_{br}} &= {\alpha _{br}}{l_{br}}\left( {{T_f} - {T_i}} \right)\\\left( {{T_f} - {T_i}} \right) &= \frac{{\Delta {l_{br}}}}{{{\alpha _{br}}{l_{br}}}}\\{T_f} &= {T_i} + \frac{{\Delta {l_{br}}}}{{{\alpha _{br}}{l_{br}}}}.\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}{T_f} &= 25^\circ {\rm{C}} + \frac{{0.015}}{{19 \times {{10}^{ - 6}}\;{\rm{/^\circ C}}}}\\ &= 25^\circ {\rm{C}} + 789.47^\circ {\rm{C}}\\ &= 814.47^\circ {\rm{C}}\end{aligned}\)

Thus, the final temperature of the brass rod is \(814.47^\circ {\rm{C}}\).