Q. 1.8 - An Introduction to Thermal Physics

Question

For a solid, we also define the linear thermal expansion coefficient, α, as the fractional increase in length per degree:

$\alpha \equiv \frac{\Delta L/L}{\Delta T}$
(a) For steel, α is 1.1 x 10^-5 K^-1. Estimate the total variation in length of a 1 km steel bridge between a cold winter night and a hot summer day.
(b) The dial thermometer in Figure 1.2 uses a coiled metal strip made of two different metals laminated together. Explain how this works.
(c) Prove that the volume thermal expansion coefficient of a solid is equal to the sum of its linear expansion coefficients in the three directions β=α_x + α_y + α_z. (So for an isotropic solid, which expands the same in all directions, β =3 α .)

Step-by-Step Solution

Answer

a) The total variation in length =0.44m

b) The coil with two metals with different value of α makes the dial thermometer to read the temperature easier.

c) The relationship $\beta =\left({\alpha }_x+{\alpha }_y+{\alpha }_z\right)$ is proved.

Part(a)Step1: Given information

coefficient of thermal expansion is α = 1.1 x 10^-5 K^-1and 

length of the steel bridge is L=1 km =1 x 10³ m.

Part(a) Step2: Explanation

Coefficient of thermal expansion of solid is given as

$\alpha =\frac{\left(\frac{\Delta L}{L}\right)}{\Delta T}$

So we can say change in length is given as 

$\Delta L=\alpha \times L\times \Delta T ......................\left(1\right)$

Lets assume the difference between cold winter temperature and hot day temperature is 40K

Substitute the values in the equation (1) we get 

$$\begin{gathered} \Delta L=\alpha \times L\times \Delta T \\ \Delta L=(1.1\times {10}^{-5} K{-1}^{})\times (1\times {10}^{3}m)\times 40 K \\ \Delta L=0.44 \mathrm{m} \end{gathered}$$

So the change in length is 0.44 m .

Part(b)Step1: Given information

A dial thermometer with two metal strips with different value of α 

Part(b)Step2: Explanation

A typical dial thermometer consists of two metal strip coils together with different values of α.

Metal with different value of α will expand differently with change of temperature.

So the coil will make a radial change by changing the temperature. 

It is easier to notice the change and hence easier to measure the temperature.

Part(c)Step1: Given information

The relationship is given 

$\beta ={\alpha }_x+{\alpha }_y+{\alpha }_z$

Prove the relationship

Part(c)Step2: Explanation

For a non-isotropic solid, they will have different α  values, i.e.,α_x , α_y , α_z

Which can be defined as, which are Coefficients of Linear expansion in all three directions are

$$\begin{gathered} {\alpha }_x=\frac{\Delta x}{x\Delta T}, \\ {\alpha }_y=\frac{\Delta y}{y\Delta T}, \\ {\alpha }_z=\frac{\Delta z}{z\Delta T} \end{gathered}$$

Where x,y and z is dimension of solid cube and Δx, Δy and Δz are changes in x, y and z respectively.

Coefficients of volume expansion is given by

$\beta =\frac{\Delta V}{V\Delta T}$

Volume of rectangular solid is 

V =  xyz .......................................(1)

Differentiate this equation 

$$\begin{gathered} \Delta V=yz\Delta x+x\Delta \left(yz\right) \\ \Delta V=yz\Delta x+xz\Delta y+xz\Delta y ...........................\left(2\right) \end{gathered}$$

Divide equation (2) by V=x y z on both the side, we get 

$\frac{\Delta V}{V}=\frac{\Delta x}{x}+\frac{\Delta y}{y}+\frac{\Delta z}{z}..............................\left(3\right)$

We know ${\alpha }_x\Delta T=\frac{\Delta x}{x},{\alpha }_y\Delta T=\frac{\Delta y}{y},{\alpha }_z\Delta T=\frac{\Delta z}{z}$

Substitute values in equation (3) ${\alpha }_x\Delta T=\frac{\Delta x}{x}, {\alpha }_y\Delta T=\frac{\Delta y}{y} and {\alpha }_z\Delta T=\frac{\Delta z}{z}$

$$\begin{gathered} \frac{\Delta V}{V}={\alpha }_x\Delta T+{\alpha }_y\Delta T+{\alpha }_z\Delta T \\ \frac{\Delta V}{V}=\left({\alpha }_x+{\alpha }_y+{\alpha }_z\right)\Delta T \\ \frac{\Delta V}{V\Delta T}=\left({\alpha }_x+{\alpha }_y+{\alpha }_z\right)...............................\left(4\right) \end{gathered}$$

We know 

$\beta =\frac{\Delta V}{V\Delta T}............................\left(5\right)$

From equation (4) and (5) we can conclude that 

$\beta =\left({\alpha }_x+{\alpha }_y+{\alpha }_z\right)$