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Question

Calculate the increase in length of a zinc rod 50.0 m50.0 \mathrm{~m} long at 15.0°C15.0{\degree} \mathrm{C} when it is heated to 130.0°C130.0{\degree} \mathrm{C}.

Solution

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Answered 2 years ago
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Given:\textbf{Given:}

l=50 ml=50 \text{ m}

T1=15CT_1=15 ^\circ \text{C}

T2=130CT_2=130 ^\circ \text{C}

α=2.6×105 C\alpha= 2.6 \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

By including the given data in the formula we get:

Δl=αl(T2T1)=2.6×10550(13015)Δl=0.15 m\begin{align*} \Delta l&=\alpha \cdot l \cdot (T_2-T_1) \\ &=2.6 \times 10^{-5} \cdot 50 \cdot (130-15)\\ &\boxed{\Delta l=0.15 \text{ m} } \end{align*}

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