HEAT EXPANSION & CONTRACTION What changes in dimensions occur when heat is extracted or added to a system ?

HEAT : EXPANSION & CONTRACTION §17.4 p576 §18.2 p617 Linear, Area, Volume Thermal expansion of water Thermal Stress (no) Molecular properties of matter References: University Physics 12th ed Young & Freedman

How does a change in temperature affect the dimensions of a system? Give examples where you have to consider the changes in the dimensions of a system when heat is added or extracted

A iron disc with a hole in it is heated. Will the diameter of the hole (a) increase, (b) decrease or (c) not change? Q

Holes get bigger

T1 < T2 Q As metal expands, the distance between any two points increases. A hole expands just as if it’s made of the same material as the hole.

A nut is very tight on a screw A nut is very tight on a screw. Which of the following is most likely to free it? (a) Cooling it (b) Heating it (c) Either (d) Neither

Bimetallic strips Two strips of different metals welded together at one temperature become more or less curved at other temperatures because the metals have different values for their coefficient of linear expansion . They are often used as thermometers and thermostats Q lower metal expands more than upper metal when heated

Most solids and liquids expand when heated. Why? Average distance between atoms Inter-atomic forces  “springs” Internal Energy U is associated with the amplitude of the oscillation of the atoms

Collisions of thermally oscillating atoms make them shift further apart Repulsive force PE Attractive force Solid heated  increased vibration of atoms  increase max displacement either side of equilibrium position  vibration is asymmetric  mean distance increases with increasing temperature E3 E2 E1 Separation of atoms THERMAL EXPANSION average distance between atoms

LINEAR THERMAL EXPANSION Ceramics (deep PE troughs) low expansion coefficients  ~10-6 K-1 Polymers high expansion coefficients  ~ 10-4 K-1 Metals  ~ 10-5 K-1  coefficient of linear expansion

Ao Lo Vo DL L A V Linear Area Volume * Simple model: assume  and  are independent of temperature, T < 100 oC * Wood expands differently in different directions

Volume expansion – solid cube Vo = Lo3 V = L3 = (Lo +  Lo T)3 = Lo3(1 +  T)3 V = Lo3 (1 + 3  T + 3 2 T2 + 3 T3) V = Lo3 (1 + 3  T) (ignoring higher order terms) V - Vo = V = 3  Lo3T =  Vo T   V0 coefficient of linear expansion  coefficient of volume expansion

Water has an anomalous coefficient of volume expansion,  is negative between 0 °C and 4 °C. Liquid water is one of the few substances with a negative coefficient of volume expansion at some temperatures (glass bottles filled with water explode in a freezer) – it does not behave like other liquids T > 4 °C water expands as temperature increases 0 < T < 4 °C water expands as temperature drops from 4 °C to 0 °C T = 3.98 °C water has its maximum density

 kg.m-3

BUOYANCY - FLOATING AND SINKING Why do ice cubes float on water?

Lakes freeze from top down rather from bottom up Water on surface cools towards 0 °C due to surrounding environment. Water as it cools and becomes more dense, it sinks carrying oxygen with it (it is most dense at about 4 °C). Warmer water moves up from below. This mixing continues until the temperature reaches 4 °C. Water then freezes first at the surface and the ice remains on the surface since ice is less dense than water (0.917 g/mL). The water at the bottom remains at 4 °C until almost the whole body of water is frozen. Without this peculiar but wonderful property of water, life on this planet may not have been possible because the body of water would have frozen from bottom up destroying all animal and plant life.

Problem B.1 As a result of a temperature rise of 32 °C a bar with a crack at its centre buckles upward. If the fixed distance between the ends of the bar is 3.77 m and the coefficient of linear expansion of the bar is 2.5x10-5 K-1, find the rise at the centre.

2Lo h = ? m Solution L L L h Lo T = 32 °C = 2.510-5 K-1 Identify / Setup T = 32 °C = 2.510-5 K-1 Lo = 3.77/2 m = 1.885 m h = ? m L = ? m Linear expansion L = Lo + L = Lo +  Lo T 2Lo L L h = ? m L h Lo

From Pythagoras’ theorem L2 = Lo2 + h2 h2 = L2 – Lo2 = (Lo +  Lo T)2 – Lo2 = 2  Lo2T + 2 Lo2 T2 h = (2  T)½ Lo neglecting very small terms h = {(2)(2.510-5)(32)}½ (1.885) m h = 0.075 m

Problem B.2 When should you buy your gas? 2 am 2 pm

When should you buy your gas? 2 am 2 pm

D depends on angle being acute or obtuse Problem B.3 A square is cut out of a copper sheet. Two straight scratches on the surface of the square intersect forming an angle   . The square is heated uniformly. As a result, the angle between the scratches A increases B decreases C stays the same D depends on angle being acute or obtuse θ

Problem B.4 A surveyor uses a steel measuring tape that is exactly 50.000 m at a temperature of 20 oC. (a) What is the length on a hot summer day when the temperature is 35 oC? (b) On the hot day the surveyor measures a distance off the tape as 35.794 m. What is the actual distance? Y & F Examples 17.2 /3. steel = 1.210-5 K-1

Part (b) is “tricky” expansion by a factor 2 Solution L0 = 50 .000 m T = 15 oC  = 1.210-5 K-1 L = L0(1 +  T) = 50.009 m Part (b) is “tricky” expansion by a factor 2 The actual distance is larger than the distance read off the tape by a factor L / L0 true distance = (35.794) (50.0009) / (50.000) m = 35.800 m