Respuesta :
Given,
The initial length of the lead pipe, L=50 m
The initial temperature, T₁=16 °C=289.15 K
The temperature of the pipe when the hot water flows through it, T₂=80 °C=353.15 K
The coefficient of linear expansion of the lead is α=29×10⁻⁶ K⁻¹
The expansion in the length of the pipe is given by,
[tex]\begin{gathered} \Delta L=\alpha\Delta T\times L \\ =\alpha\times(T_2-T_1)L \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} \Delta L=29\times10^{-6}\times(353.15-289.15)\times50 \\ =0.093\text{ m} \end{gathered}[/tex]Therefore the pipe expands by a length of 0.093 m
Thus the new length of the pipe is
[tex]\begin{gathered} L_n=L+\Delta L \\ =50+0.093 \\ =50.093\text{ m} \end{gathered}[/tex]Thus the length of the hot pipe is 50.093 m
