Question

A brass rod and an iron rod differ in length by 28cm at 20°C. What should be the original length of the iron rod for the difference in length to remain the same when both rods are heated to 90°C? Linear expansivity of brass=1.9×10^-5. Linear expansivity of iron=1.2×10^-5.

Pls show workings.

Answer 1

Answer:

The original length of the iron rod is approximately 572.189 meters.

Explanation:

This is a case of linear expansion, which is defined by the following differential equation:

$\alpha = \frac{1}{L}\cdot \frac{dL}{dT}$ (1)

Where:

$\alpha$ - Linear expansion coefficient, measured in $\frac{1}{^{\circ}C}$.

$L$ - Length of the element, measured in centimeters.

$\frac{dL}{dT}$ - First derivative of the length of the element with respect to temperature, measured in centimeters per degree Celsius.

If we assume that thermal deformation are small regarding the length of the element, then we simplify (1) in the following form:

$\alpha \approx \frac{\Delta L_{o}}{L\cdot \Delta T}$

$L_{f} -L_{o} = \alpha \cdot L_{o} \cdot \Delta T$

$L_{f} = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})]$ (2)

Where:

$L_{o}$, $L_{f}$ - Initial and final lengths of the element, measured in centimeters.

$T_{o}, T_{f}$ - Initial and final temperatures of the element, measured in degrees Celsius.

Given that brass has a higher coefficient of linear expansion, it is suppose that initial length is less than the initial length of the iron element. Then, we have the following system of linear equations:

Brass

$L = L_{o}\cdot [1+\alpha_{B}\cdot (T_{f}-T_{o})]$ (3)

Iron

$L = (L_{o}+28)\cdot [1+\alpha_{I}\cdot (T_{f}-T_{o})]$ (4)

Where $\alpha_{B}$, $\alpha_{I}$ are coefficients of linear expansion of brass and iron, measured in $\frac{1}{^{\circ}C}$.

By equalizing (3) and (4), we have the following formula:

$L_{o} \cdot [1+\alpha_{B}\cdot (T_{f}-T_{o})] = (L_{o}+28)\cdot [1+\alpha_{I}\cdot (T_{f}-T_{o})]$

$L_{o} \cdot (\alpha_{B}-\alpha_{I})\cdot (T_{f}-T_{o}) = 28\cdot [1+\alpha_{I}\cdot (T_{f}-T_{o})]$

$L_{o} = \frac{28\cdot [1+\alpha_{I}\cdot (T_{f}-T_{o})]}{(\alpha_{B}-\alpha_{I})\cdot (T_{f}-T_{o} )}$

If we know that $\alpha_{B} = 1.9\times 10^{-5}\,\frac{1}{^{\circ}C}$, $\alpha_{I} = 1.2\times 10^{-5}\,\frac{1}{^{\circ}C}$, $T_{o} = 20\,^{\circ}C$ and $T_{f} = 90\,^{\circ}C$, then the initial length of the iron rod is:

$L_{o} = \frac{28\cdot [1+\left(1.2\times 10^{-5}\,\frac{1}{^{\circ}C} \right)\cdot (90\,^{\circ}C-20\,^{\circ}C)]}{\left(1.9\times 10^{-5}\,\frac{1}{^{\circ}C}-1.2\times 10^{-5}\,\frac{1}{^{\circ}C} \right)\cdot (90\,^{\circ}C-20\,^{\circ}C)}$

$L_{o} = 57190.857\,cm$

$L_{o, I} = 57218.857\,cm$

$L_{o,I} = 572.189\,m$

The original length of the iron rod is approximately 572.189 meters.