## Related questions with answers

A brass rod $1.020 \mathrm{~m}$ long expands $3.0 \mathrm{~mm}$ when it is heated. Calculate the temperature change.

Solution

Verified$\textbf{Given:}$

$l=1.02 \text{ m}$

$\Delta l= 3 \text { mm}$

$\alpha= 1.9 \times 10^{-5} \dfrac{\text{ }}{^\circ\text{C}}$

The change in the length of the material under the influence of the change in temperature is given by the formula:

$\Delta l=\alpha \cdot l \cdot \Delta T$

From the formula we express $\Delta T$:

$\Delta T=\dfrac{ \Delta l }{\alpha \cdot l}$

We first express $\Delta l$ in meters using $1 \text { m} = 1000 \text { mm}$

$\Delta l= 3 \text { mm}=0.003 \text{ m}$

By including the given data in the formula we get:

$\begin{align*} \Delta T&=\dfrac{ 0.003}{1.9 \times 10^{-5} \cdot 1.02}\\ &\boxed{\Delta T=15.5 ^\circ \text{C} } \end{align*}$

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