Solved Problem on Thermal Expansion

The average linear expansion coefficient of iron is equal to 0.0000117 °C⁻¹. By how much should the temperature of an iron block be increased so that its volume increases by 1%?

Problem data:

Linear expansion coefficient of iron: α = 0.0000117 °C⁻¹;
Change in volume: ΔV = 1%.

Problem diagram:

Figure 1

Solution:

The volume variation is 1%

\[ \begin{gather} \Delta V=1V_0 \\[5pt] \Delta V=\frac{1}{100}V_0 \\[5pt] \Delta V=0.01V_0 \end{gather} \]

The problem provides us with the linear expansion coefficient of the material. To calculate the volume increase, we need the volumetric expansion coefficient, which will be

\[ \begin{gather} \bbox[#99CCFF,10px] {\gamma=3\alpha} \end{gather} \]

\[ \begin{gather} \gamma=3\times 0.0000117\;\mathrm{°C^{-1}} \\[5pt] \gamma=0.0000351\;\mathrm{°C^{-1}} \\[5pt] \gamma=3.51\times 10^{-5}\;\mathrm{°C^{-1}} \end{gather} \]

The final volume is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {\Delta V=\gamma V_0\Delta t} \end{gather} \]

substituting the problem data, we find the temperature variation

\[ \begin{gather} 0.01\cancel{V_0}=\left(3.51\times 10^{-5}\;\mathrm{°C^{-1}}\right)\cancel{V_0}\Delta t \\[5pt] 0,01=\left(3.51\times 10^{-5}\;\mathrm{°C^{-1}}\right)\Delta t \\[5pt] \Delta t=\frac{0.01}{3.51\times 10^{-5}\;\mathrm{°C^{-1}}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\Delta t\approx 285\;\mathrm{°C}} \end{gather} \]

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