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

To calculate the original length of the iron rod needed to maintain a 28cm difference in length when heated to 90°C, set the linear expansion of both rods equal to each other using their respective expansivity coefficients and solve for the iron rod's length, which is found to be 350 cm.

Step-by-step explanation:

The student is asking about the linear thermal expansion that occurs when materials are heated and they want to know how to maintain a constant difference in length between a brass rod and an iron rod as they are heated. To find the original length of the iron rod, one must first calculate the change in length for each material using the formula ΔL = αLΔT, where ΔL is the change in length, α is the coefficient of linear expansion, L is the original length, and ΔT is the change in temperature. However, since we want the difference in length to remain the same, the expansion of the iron rod must equal the expansion of the brass rod.

To calculate, let the original length of the iron rod be L. The change in length for the iron rod is 1.2×10^-5 × L × (90 °C - 20 °C), and for the brass rod, it's 1.9×10^-5 × (L+28 cm) × (90 °C - 20 °C). Since their expansions must be equal, we can set these two expressions equal to each other and solve for L. Doing so gives us: L = 280 cm / (1.2×10^-5 - 1.9×10^-5) = 350 cm as the original length of the iron rod so that when heated to 90°C, the difference in length remains 28 cm.

Answer 2

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

Step-by-step explanation:

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

`\alpha = (1)/(L)\cdot (dL)/(dT)` (1)

Where:

`\alpha` - Linear expansion coefficient, measured in
`(1)/(^(\circ)C)`.

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

`(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 (\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
`(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) = (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* 10^(-5)\,(1)/(^(\circ)C)`,
`\alpha_(I) = 1.2* 10^(-5)\,(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) = (28\cdot [1+\left(1.2* 10^(-5)\,(1)/(^(\circ)C) \right)\cdot (90\,^(\circ)C-20\,^(\circ)C)])/(\left(1.9* 10^(-5)\,(1)/(^(\circ)C)-1.2* 10^(-5)\,(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.