Question

A metal rod has a length of 50cm at $20{}^\circ C$. When it is heated to $95{}^\circ C$, its length becomes 50.06 cm. Find the length of rod at $0{}^\circ C$.

Answer 1

The change in length due to change is the change in temperature is called Linear Thermal Expansion. The change is proportional to the length of the substance. It also depends upon the coefficient of linear expansion if the material. The relation of these factors can be formalised and has been given below.
Formula Used:
$\alpha =\dfrac{\Delta L}{{{L}_{1}}\Delta T}$
$\Delta L={{L}_{2}}-{{L}_{1}}$
$\Delta T={{T}_{2}}-{{T}_{1}}$
${{L}_{3}}={{L}_{1}}(1+\alpha \cdot \Delta T')$

Complete answer:
For solving the question, we have to realise that the metal rod is changing its length due to thermal expansion and contraction. So, we start off by calculating the coefficient of linear expansion ($\alpha$) for the metal rod. The coefficient of linear expansion ($\alpha$) is given by
$\alpha =\dfrac{\Delta L}{{{L}_{1}}\Delta T}$
To calculate the various elements of the equation, let’s use the information given in the question.
According to the question.

& {{L}_{1}}=50cm \\\ & {{L}_{2}}=50.06 \\\ \end{aligned}$$ So, $$\begin{aligned} & \Delta L={{L}_{2}}-{{L}_{1}} \\\ & \Delta L=50.06-50 \\\ & \Delta L=0.06cm \\\ \end{aligned}$$ Now for the change in temperature $$\Delta T={{T}_{2}}-{{T}_{1}}$$ $$\begin{aligned} & {{T}_{1}}=20{}^\circ C \\\ & {{T}_{2}}=95{}^\circ C \\\ \end{aligned}$$ $$\begin{aligned} & \Delta T={{T}_{2}}-{{T}_{1}} \\\ & \Delta T=95-20 \\\ & \Delta T=75{}^\circ C \\\ \end{aligned}$$ Now that we have all the elements let’s calculate the coefficient of linear expansion ($$\alpha $$) Again, $$\alpha =\dfrac{\Delta L}{{{L}_{1}}\Delta T}$$ $$\begin{aligned} & \alpha =\dfrac{0.06}{50\times 75} \\\ & \alpha =16\times {{10}^{-6}}{}^\circ {{C}^{-1}} \\\ \end{aligned}$$ And now we have the coefficient of linear expansion ($$\alpha $$). To find out the length of the metal rod at $$0{}^\circ C$$we will have to consider another change in length of the metal rod. Let us consider the change in due in length due to falling of temperature from $$20{}^\circ C$$to $$0{}^\circ C$$ Let $${{L}_{3}}$$be the length of rod at $$0{}^\circ C$$ Now, $${{T}_{1}}=20{}^\circ C$$ and $${{T}_{3}}=0{}^\circ C$$, thus $$\begin{aligned} & \Delta T'={{T}_{3}}-{{T}_{1}} \\\ & \Delta T'=0-20 \\\ & \Delta T'=-20 \\\ \end{aligned}$$ We also know that the new changed after thermal expansion can be found by $${{L}_{3}}={{L}_{1}}(1+\alpha \cdot \Delta T')$$ Putting in all the values as calculated above, $$\begin{aligned} & {{L}_{3}}={{L}_{1}}(1+\alpha \cdot \Delta T') \\\ & {{L}_{3}}=50(1+16\times {{10}^{-6}}\cdot (-20)) \\\ & {{L}_{3}}=49.984cm \\\ \end{aligned}$$ So, the length of the rod at $$0{}^\circ C$$ would be 49.984 cm. **Note:** The coefficient of linear expansion ($$\alpha $$) is an intrinsic property of a compound or a substance. Each substance has its own coefficient of linear expansion. Solids have higher coefficients of linear expansion as the cohesive interatomic forces are strongest in solids.