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Q. 1.7

Expert-verified
Found in: Page 6

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 550,000 . The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β :$\beta \equiv \frac{\Delta V/V}{\Delta T}$(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β =1 / 550,000 K-1=1.81 x 10-4 K-1. (The exact value varies with temperature, but between 0oC and 200oC the variation is less than 1 %.)(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 x 10 -4 K-1 at 100oC, but decreases as the temperature is lowered until it becomes zero at 4oC. Below 4oC it is slightly negative, reaching a value of -0.68 x 10-4K-1 at 0oC. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.

a) Diameter of the bulb is = 2 .8 x 10-3m

b) Water has very different behavior of thermal expansion.

See the step by step solution

Part(a) Step1: Given information

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

Part(a)Step2: Explanation

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

Volume can be calculate by using

V = π r2h

Substitute the values we get

$V=\pi {\left(0.2×{10}^{-2}m\right)}^{2}×\left(1×{10}^{-2}m\right)\phantom{\rule{0ex}{0ex}}V=1.25×{10}^{-7}{\mathrm{m}}^{3}$

Coefficient of thermal expansion is calculated as
$\beta =\frac{\Delta V}{V\Delta T}\phantom{\rule{0ex}{0ex}}Simplify,\phantom{\rule{0ex}{0ex}}\Delta V=\beta V\Delta T$

Substitute the value in above equation to calculate change in volume

$\Delta V=\left(1.81×{10}^{-4}{K}^{-1}\right)×\left(1.256×{10}^{-7}{m}^{3}\right)×\left(274.15K\right)\phantom{\rule{0ex}{0ex}}\Delta V=6.19×{10}^{-9}{\mathrm{m}}^{3}$

The inner radius of the thermometer bulb ri can be obtained from the change in volume
$\Delta V=\pi {{r}_{i}}^{2}\Delta l\phantom{\rule{0ex}{0ex}}Solveforradius\phantom{\rule{0ex}{0ex}}{r}_{i}=\sqrt{\frac{\Delta V}{\pi \Delta l}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Substitutevalues\phantom{\rule{0ex}{0ex}}{r}_{i}=\sqrt{\frac{6.19×{10}^{-9}{m}^{3}}{\pi ×{10}^{-3}m}}\phantom{\rule{0ex}{0ex}}{r}_{i}=1.4×{10}^{-3}\mathrm{m}$

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

Part(B)Step1: Given information

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

Part(b) Step2: Explanation

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 0oC and 4oC i.e., ice generally seems to contract as temperature goes up to 4oC.

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.

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