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Physics / Physics: Principles with Applications 6 / Chapter 13 / Problem 21P

Physics: Principles with Applications | 6th Edition | ISBN: 9780321569837 | Authors: Douglas C. Giancoli

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(a) The tube of a mercury thermometer has an inside diameter of 0.140 mm. The bulb has a

Chapter 13, Problem 21P

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QUESTION:

Problem 21P

(a) The tube of a mercury thermometer has an inside diameter of 0.140 mm. The bulb has a volume of 0.255 cm3. How far will the thread of mercury move when the temperature changes from 11.5°C to 33.0°C? Take into account expansion of the Pyrex glass. (b) Determine a formula for the change in length of the mercury column in terms of relevant variables. Ignore tube volume compared to bulb volume.

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QUESTION:

Problem 21P

(a) The tube of a mercury thermometer has an inside diameter of 0.140 mm. The bulb has a volume of 0.255 cm3. How far will the thread of mercury move when the temperature changes from 11.5°C to 33.0°C? Take into account expansion of the Pyrex glass. (b) Determine a formula for the change in length of the mercury column in terms of relevant variables. Ignore tube volume compared to bulb volume.

ANSWER:

Solution 21P

Step 1 of 2

Part a

We are required to calculate the distance moved by mercury on increasing the temperature.

The diameter of the mercury tube $=0.140$ mm

The radius of the mercury tube $r=0.070$ mm

$r=0.070\times{}1{0}^{-3}$ m

The area of the tube, $A=\pi{}{r}^{2}$

$A=3.14\times{}(0.070\times{}1{0}^{-3}{)}^{2}$ m2

$A=1.53\times{}1{0}^{-8}$ m2

Let the distance moved by the mercury is $x$ .

So, the volume change is $=Ax=1.53\times{}1{0}^{-8}x$ m3 …..(1)

The initial volume of mercury $V{\ }_{0,mercury}=0.255$ cm3 $=0.255\times{}1{0}^{-6}$ m3

The temperature change is $\Delta{}T=33.0-11.5$ 0C

$\Delta{}T=21.{5}^{0}$ C

The coefficient of volume expansion of mercury is $\beta{}{\ }_{mercury}=180\times{}1{0}^{-6}$ (C0)-1

The change in volume of mercury is,

$\Delta{}V{\ }_{mercury}=V{\ }_{0,mercury}\times{}\beta{}{\ }_{mercury}\times{}\Delta{}T$

$\Delta{}V{\ }_{mercury}=0.255\times{}1{0}^{-6}\times{}180\times{}1{0}^{-6}\times{}21.{5}^{0}$ m3

$\Delta{}V{\ }_{mercury}=986.85\times{}1{0}^{-12}$ m3

The coefficient of volume expansion of pyrex is $\beta{}{\ }_{pyrex}=9\times{}1{0}^{-6}$ (C0)-1

The change in volume of pyrex is,

$\Delta{}V{\ }_{pyrex}=V{\ }_{0,mercury}\times{}\beta{}{\ }_{pyrex}\times{}\Delta{}T$

$\Delta{}V{\ }_{pyrex}=0.255\times{}1{0}^{-6}\times{}9\times{}1{0}^{-6}\times{}21.{5}^{0}$ m3

$\Delta{}V{\ }_{pyrex}=49.340\times{}1{0}^{-12}$ m3

Therefore, the overall volume change $=\Delta{}V{\ }_{mercury}-\Delta{}V{\ }_{pyrex}=937.51\times{}1{0}^{-12}$ m3…..(2)

From equations (1) and (2),

$1.53\times{}1{0}^{-8}x=937.51\times{}1{0}^{-12}$

$x=612\times{}1{0}^{-4}$ m

$x=6.12$ cm

Therefore, the mercury thread will move by 6.12 cm.

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