A brass rod 1.020 m1.020 \mathrm{~m} long expands 3.0 mm3.0 \mathrm{~mm} when it is heated. Calculate the temperature change.


Answered 2 years ago
Answered 2 years ago
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l=1.02 ml=1.02 \text{ m}

Δl=3 mm\Delta l= 3 \text { mm}

α=1.9×105 C\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:

Δl=αlΔT\Delta l=\alpha \cdot l \cdot \Delta T

From the formula we express ΔT\Delta T:

ΔT=Δlαl\Delta T=\dfrac{ \Delta l }{\alpha \cdot l}

We first express Δl\Delta l in meters using 1 m=1000 mm1 \text { m} = 1000 \text { mm}

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

By including the given data in the formula we get:

ΔT=0.0031.9×1051.02ΔT=15.5C\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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