91Ó°ÊÓ

for mercury, β =1 / 550,000 K^-1=1.81 x 10^-4 K^-1.

Lets assume that a typical mercury thermometer with cylindrical bulb having
height h= 1 cm = 1 x 10^-2 m
radius r=0.2 cm = 0.2 x 10^-2 m
and the scale on the thermometer is 1 mm (= 1 x 10^-3 m) per degree.

Volume can be calculate by using

V = Ï€ r²h

Substitute the values we get

$$\begin{gathered} V=\mathrm{π}{\left(0.2×{10}^{-2}m\right)}^2×\left(1×{10}^{-2}m\right) \\ V=1.25×{10}^{-7}{\mathrm{m}}^3 \end{gathered}$$

Coefficient of thermal expansion is calculated as
$$\begin{gathered} \mathrm{β}=\frac{\mathrm{Δ}V}{V\mathrm{Δ}T} \\ Simplify, \\ \mathrm{Δ}V=\mathrm{β}V\mathrm{Δ}T \end{gathered}$$

Substitute the value in above equation to calculate change in volume

$$\begin{gathered} \mathrm{Δ}V=(1.81×{10}^{-4}{K}^{-1})×(1.256×{10}^{-7}{m}^3)×(274.15K) \\ \mathrm{Δ}V=6.19×{10}^{-9}{\mathrm{m}}^3 \end{gathered}$$

The inner radius of the thermometer bulb r_i can be obtained from the change in volume
$$\begin{gathered} \mathrm{Δ}V=\mathrm{π}{r_i}^2\mathrm{Δ}l \\ Solveforradius \\ {r}_i=\sqrt{\frac{\mathrm{Δ}V}{\mathrm{π}\mathrm{Δ}l}} \\ Substitutevalues \\ {r}_i=\sqrt{\frac{6.19×{10}^{-9}{m}^3}{\mathrm{π}×{10}^{-3}m}} \\ {r}_i=1.4×{10}^{-3}\mathrm{m} \end{gathered}$$

So diameter is 2 x radius = 2 x 1.4 x 10^-3m = 2 .8 x 10^-3m

The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 x 10 ^-4 K^-1 at 100^oC, but decreases as the temperature is lowered until it becomes zero at 4^oC. Below 4^oC it is slightly negative, reaching a value of -0.68 x 10^-4K^-1 at 0^oC.

Thermal expansion coefficient of water behaves very differently.

β varies much in the liquid region.

As the temperature drops it decreases. But value of β is negative in between 0^oC and 4^oC i.e., ice generally seems to contract as temperature goes up to 4^oC.

If β were positive over the entire region of range , means always in one side of the curve.

This would lead to a water body (Lake, Pond) would start to freeze from bottom up Not from from the top down.