Question

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as the drawing shows, two thin strips of metal are bolted together at one end and have the same temperature. one is steel, and the other is aluminum. the steel strip is 0.11% longer than the aluminum strip. by how much should the temperature of the strips be increased, so that the strips have the same length?
steel
aluminum
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units

Explanation:

Step1: Write the formula for linear expansion

The formula for linear expansion is \( \Delta L = L_0\alpha\Delta T \), where \( \Delta L \) is the change in length, \( L_0 \) is the original length, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature. Let the original length of the aluminum strip be \( L_{0,Al} = L \), then the original length of the steel strip is \( L_{0,Steel}=L(1 + 0.0011)\). The coefficient of linear expansion for aluminum \( \alpha_{Al}=23\times10^{-6}\ ^{\circ}C^{-1}\) and for steel \( \alpha_{Steel} = 12\times10^{-6}\ ^{\circ}C^{-1}\). After heating, \( L_{0,Al}(1+\alpha_{Al}\Delta T)=L_{0,Steel}(1+\alpha_{Steel}\Delta T)\)

Step2: Substitute the values into the equation

Substitute \( L_{0,Al} = L \) and \( L_{0,Steel}=L(1 + 0.0011)\) into \( L_{0,Al}(1+\alpha_{Al}\Delta T)=L_{0,Steel}(1+\alpha_{Steel}\Delta T)\). We get \( L(1+\alpha_{Al}\Delta T)=L(1 + 0.0011)(1+\alpha_{Steel}\Delta T)\). Divide both sides by \( L \): \(1+\alpha_{Al}\Delta T=(1 + 0.0011)(1+\alpha_{Steel}\Delta T)\). Expand the right - hand side: \(1+\alpha_{Al}\Delta T=1+0.0011+\alpha_{Steel}\Delta T+0.0011\alpha_{Steel}\Delta T\). Since \(0.0011\alpha_{Steel}\Delta T\) is a very small term (because \(0.0011\) and \( \alpha_{Steel}\) are both small), we can neglect it. So, \( \alpha_{Al}\Delta T=0.0011+\alpha_{Steel}\Delta T\)

Step3: Solve for \(\Delta T\)

Rearrange the equation \( \alpha_{Al}\Delta T-\alpha_{Steel}\Delta T=0.0011\). Factor out \( \Delta T\): \( \Delta T(\alpha_{Al}-\alpha_{Steel})=0.0011\). Then \( \Delta T=\frac{0.0011}{\alpha_{Al}-\alpha_{Steel}}\). Substitute \( \alpha_{Al}=23\times10^{-6}\ ^{\circ}C^{-1}\) and \( \alpha_{Steel} = 12\times10^{-6}\ ^{\circ}C^{-1}\) into the formula: \( \Delta T=\frac{0.0011}{(23 - 12)\times10^{-6}\ ^{\circ}C^{-1}}\)

Step4: Calculate the value of \(\Delta T\)

\( \Delta T=\frac{0.0011}{11\times10^{-6}\ ^{\circ}C^{-1}}=\frac{1.1\times10^{-3}}{11\times10^{-6}\ ^{\circ}C^{-1}} = 100^{\circ}C\)

Answer:

\(100^{\circ}C\)