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,
![\begin{gathered} \Delta L=\alpha\Delta T* L \\ =\alpha*(T_2-T_1)L \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/50txe3dlojo2o832h4xfx3002vovbz9if6.png)
On substituting the known values,
![\begin{gathered} \Delta L=29*10^(-6)*(353.15-289.15)*50 \\ =0.093\text{ m} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/dopwnr60a8bvng1wwhmpxdnvzsobhk4hif.png)
Therefore the pipe expands by a length of 0.093 m
Thus the new length of the pipe is
![\begin{gathered} L_n=L+\Delta L \\ =50+0.093 \\ =50.093\text{ m} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/pq7cumhkzk7c74hy1w8qmu1v35h02gq63z.png)
Thus the length of the hot pipe is 50.093 m