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Question: At \({15^o}C\) the height of the Eiffel tower is \(324\,m\). If it is made of iron, what will be the...

At 15oC{15^o}C the height of the Eiffel tower is 324m324\,m. If it is made of iron, what will be the increase in length in cm, at 30oC{30^o}C?

Explanation

Solution

Thermal expansion, which does not include phase transitions, is the tendency of matter to alter its form, area, volume, and density in response to a change in temperature. The average molecular kinetic energy of a material is a monotonic function of temperature. When a substance is heated, the molecules begin to vibrate and move more, resulting in a greater distance between them.

Formula used:
Δl=lαΔt\Delta l = l\alpha \Delta t
Coefficient of linear expansion = α\alpha
Increase in length = Δl\Delta l
length = ll
Δt\Delta t= The temperature is rising

Complete step by step answer:
It's difficult to find substances that contract as the temperature rises, and they only happen in a few temperature ranges. The material's coefficient of linear thermal expansion is defined as the relative expansion (also known as strain) divided by the change in temperature, and it fluctuates with temperature. Particles move faster as their energy increases, reducing the intermolecular interactions between them and thereby expanding the material.

Because thermal expansion drops as bond energy increases, which also affects the melting point of solids, materials having a high melting point are more likely to have reduced thermal expansion. Liquids expand slightly more than solids in general. When compared to crystals, glasses have a larger thermal expansion.

Rearrangements in an amorphous material at the glass transition temperature cause distinct discontinuities in coefficient of thermal expansion and specific heat. These discontinuities enable for the identification of the glass transition temperature, which is the temperature at which a supercooled liquid becomes a glass.
Using Δl=lαΔt\Delta l = l\alpha \Delta t
Coefficient of linear expansion of iron = α\alpha (Its temperature is 11.5 ×10611.5{\text{ }} \times {10^{ - 6}}degrees Celsius)
increase in length = Δl\Delta l=?
length = ll= 32400 cm
Δt\Delta t= The temperature is rising = 30oC15oC=15oC{30^o}C - {15^o}C = {15^o}C
So substituting,
Δll=α1ΔT\dfrac{{\Delta {\text{l}}}}{{\text{l}}} = {\alpha _1}\Delta {\text{T}}
So,
Δl32400=11.5×106×15\Rightarrow \dfrac{{\Delta {\text{l}}}}{{32400}} = 11.5 \times {10^{ - 6}} \times 15
Δl=32400×11.5×106×15 \Rightarrow \Delta {\text{l}} = 32400 \times 11.5 \times {10^{ - 6}} \times 15 \\\ \therefore \Delta {\text{l}} = 5.6;{\text{cm}}$

Hence, the increase in length in cm is 5.6  cm5.6\;{\text{cm}}.

Note: Make sure you calculate the change in temperature and coefficient of thermal expansion properly. The coefficient of thermal expansion indicates how an object's size varies as the temperature varies. It calculates the fractional change in size per degree change in temperature at constant pressure, with lower coefficients indicating a reduced tendency for size change. There are three different sorts of coefficients: volumetric, area, and linear. The coefficient to use is determined by the application and whatever dimensions are regarded as significant. For solids, the change along a length or over a specific region may be all that matters.