Question

A length of lead piping is 50 m long at a temperature of 16 °C. When hot waterflows through it the temperature of the pipe rises to 80 °C. Calculate the length ofthe hot pipe.

Answer 1

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}`

On substituting the known values,

`\begin{gathered} \Delta L=29*10^(-6)*(353.15-289.15)*50 \\ =0.093\text{ m} \end{gathered}`

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}`

Thus the length of the hot pipe is 50.093 m