# 1.3 Thermal expansion  (Page 6/10)

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How much taller does the Eiffel Tower become at the end of a day when the temperature has increased by $15\phantom{\rule{0.2em}{0ex}}\text{°}\text{C?}$ Its original height is 321 m and you can assume it is made of steel.

What is the change in length of a 3.00-cm-long column of mercury if its temperature changes from $37.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ to $40.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ , assuming the mercury is constrained to a cylinder but unconstrained in length? Your answer will show why thermometers contain bulbs at the bottom instead of simple columns of liquid.

Using [link] to find the coefficient of thermal expansion of mercury:
$\text{Δ}L=\alpha L\text{Δ}T=\left(6.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\text{/}\text{°}\text{C}\right)\left(0.0300\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(3.00\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}\right)=5.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{m}$ .

How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature $35.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ greater than when they were laid? Their original length is 10.0 m.

You are looking to buy a small piece of land in Hong Kong. The price is “only” $60,000 per square meter. The land title says the dimensions are $20\phantom{\rule{0.2em}{0ex}}\text{m}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}30\phantom{\rule{0.2em}{0ex}}\text{m}$ . By how much would the total price change if you measured the parcel with a steel tape measure on a day when the temperature was $20\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ above the temperature that the tape measure was designed for? The dimensions of the land do not change. On the warmer day, our tape measure will expand linearly. Therefore, each measured dimension will be smaller than the actual dimension of the land. Calling these measured dimensions $l\text{'}$ and $w\text{'}$ , we will find a new area, A . Let’s calculate these measured dimensions: $l\text{'}={l}_{0}-\text{Δ}l=\left(20\phantom{\rule{0.2em}{0ex}}\text{m}\right)-\left(20\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}\right)\left(20\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(\frac{1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}}{\text{°}\text{C}}\right)=19.9952\phantom{\rule{0.2em}{0ex}}\text{m}$ ; $A\text{'}=l\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}w\text{'}=\left(29.9928\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(19.9952\phantom{\rule{0.2em}{0ex}}\text{m}\right)=599.71\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}$ ; Cost change $=\left(A-A\text{'}\right)\left(\frac{60,000}{{\text{m}}^{2}}\right)=\left(\left(600-599.71\right){\text{m}}^{2}\right)\left(\frac{60,000}{{\text{m}}^{2}}\right)=17,000$ . Because the area gets smaller, the price of the land decreases by about$17,000.

Global warming will produce rising sea levels partly due to melting ice caps and partly due to the expansion of water as average ocean temperatures rise. To get some idea of the size of this effect, calculate the change in length of a column of water 1.00 km high for a temperature increase of $1.00\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . Assume the column is not free to expand sideways. As a model of the ocean, that is a reasonable approximation, as only parts of the ocean very close to the surface can expand sideways onto land, and only to a limited degree. As another approximation, neglect the fact that ocean warming is not uniform with depth.

(a) Suppose a meter stick made of steel and one made of aluminum are the same length at ${0}^{}\text{°}\text{C}$ . What is their difference in length at $22.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ ? (b) Repeat the calculation for two 30.0-m-long surveyor’s tapes.

a. Use [link] to find the coefficients of thermal expansion of steel and aluminum. Then $\text{Δ}{L}_{\text{Al}}-\text{Δ}{L}_{\text{steel}}=\left({\alpha }_{\text{Al}}-{\alpha }_{\text{steel}}\right){L}_{0}\text{Δ}T=\left(\frac{2.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}}{\text{°}\text{C}}-\frac{1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}}{\text{°}\text{C}}\right)\left(1.00\phantom{\rule{0.2em}{0ex}}\text{m}\right)\left(22\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}\right)=2.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{m}$ .
b. By the same method with ${L}_{0}=30.0\phantom{\rule{0.2em}{0ex}}\text{m}$ , we have $\text{Δ}L=8.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{m}$ .

(a) If a 500-mL glass beaker is filled to the brim with ethyl alcohol at a temperature of $5.00\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ , how much will overflow when the alcohol’s temperature reaches the room temperature of $22.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ ? (b) How much less water would overflow under the same conditions?

Most cars have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to its 16.0-L capacity when at $10.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . What volume of radiator fluid will overflow when the radiator and fluid reach a temperature of $95.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C,}$ given that the fluid’s volume coefficient of expansion is $\beta =400\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\text{/}\text{°}\text{C}$ ? (Your answer will be a conservative estimate, as most car radiators have operating temperatures greater than $95.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ ).

$\text{Δ}V=0.475\phantom{\rule{0.2em}{0ex}}\text{L}$

A physicist makes a cup of instant coffee and notices that, as the coffee cools, its level drops 3.00 mm in the glass cup. Show that this decrease cannot be due to thermal contraction by calculating the decrease in level if the $350\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}$ of coffee is in a 7.00-cm-diameter cup and decreases in temperature from $95.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ to $45.0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . (Most of the drop in level is actually due to escaping bubbles of air.)

The density of water at $0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ is very nearly $1000\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}$ (it is actually $999.84\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}$ ), whereas the density of ice at $0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ is $917\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}.$ Calculate the pressure necessary to keep ice from expanding when it freezes, neglecting the effect such a large pressure would have on the freezing temperature. (This problem gives you only an indication of how large the forces associated with freezing water might be.)

If we start with the freezing of water, then it would expand to $\left(1\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}\right)\left(\frac{1000\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}}{917\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}}\right)=1.09\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}=1.98\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}$ of ice.

Show that $\beta =3\alpha ,$ by calculating the infinitesimal change in volume dV of a cube with sides of length L when the temperature changes by dT .

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oky
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good
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Cyclone