Thermal Expansion

THERMAL PROPERTIES OF MATTER THERMAL EXPANSION • • When matter is heated without any change in it's state, it usually expands. According to atomic theory of matter, asymmetry in potential energy curve is responsible for thermal expansion. As with rise in temperature the amplitude of vibration increases and hence energy of atoms increases, hence the average distance between the atom increases. So the matter as a whole expands. Thermal expansion is minimum in case of solids but maximum in case of gases because intermolecular force is maximum in solids but minimum in gases. Solids can expand in one dimension (Linear expansion), two dimension (Superficial expansion) and three dimension (Volume expansion) while liquids and gases usually suffers change in volume only. Linear expansion :  = 0 (1 + )  = 0 Superficial (areal) expansion : A = A0 (1 + ) Also A0 = 02 and A = 2 So 2 = 02(1 + ) = [0(1 + )]2  =2 Volume expansion : V = V0 (1 + ) Also V = 3 and V0=03 so =3  6 = 3 = 2 or  :  :  = 1 : 2 : 3 Contraction on heating : Some rubber like substances contract on heating because transverse vibration of atoms of substance dominate over longitudinal vibration which is responsible for expansion. Application of thermal Expansion in Solids (a) Bi–metallic strip : Two strips of equal length but of different materials (different coefficient of linear expansion) when join together, it is called "Bi–metallic strip" and can be used in thermostat to break or make electrical contact. This strip has the characteristic property of bending on heating due to unequal linear expansion of the two metals. The strip will bend with metal of greater  on outer side. Steel Brass Room temperature Steel Brass Higher temperature ON At room temperature OFF Bimetallic strip At high temperature (b) Effect of temperature on the time period of a simple pendulum : A pendulum clock keeps proper time at temperature  If temperature is increased to ' ( >) then due to linear expansion, length of pendulum increases and hence its time period will increase. Fractional change in time period T 1 l T 1  ) =  (Q T  l  T 2 l T 2 (i) Due to increment in its time period, a pendulum clock becomes slow in summer and will lose time. Loss of time in a time period T= 1  T 2 (ii) The clock will lose time i.e. will become slow if '> (in summer) and will gain time i.e will become fast if ' < (in winter). (iii) Since coefficient of linear expansion () is very small for invar, hence pendulums are made of invar to show the correct time in all seasons. (c) When a rod whose ends are rigidly fixed such as to prevent expansion or contraction, undergoes a change in temperature due to thermal expansion or contraction, a compressive or tensile stress is developed in it. Due to this thermal stress the rod will exert a large force on the supports. If the change in temperature of a rod of length L is  then :– Heating Thermal strain = L ΔL 1 =  α = × L L Δθ Cooling So thermal stress = Y Y  stress strain So force on the supports F=YA (d) Error in scale reading due to expansion or contraction: If a scale gives correct reading at temperature  At temperature '(>) due to linear expansion of scale, the scale will expand and scale reading will be lesser than true value so that, (e) Expansion of cavity: Thermal expansion of an isotropic object may be imagined as a photographic enlargement. THERMAL PROPERTIES OF MATTER (f) Some other application (i) When rails are laid down on the ground, space is left between the ends of two rails (ii) The transmission cable are not tightly fixed to the poles (iii) Test tubes, beakers and cubicles are made of pyrex–glass or silica because they have very low value of coefficient of linear expansion (iv) The iron rim to be put on a cart wheel is always of slightly smaller diameter than that of wheel (v) A glass stopper jammed in the neck of a glass bottle can be taken out by warming the neck of the bottle. Ex. A steel ruler exactly 20 cm long is graduated to give correct measurements at 200C. (  steel  1.2  10 5 C 1 ) (a) Will it give readings that are too long or too short at lower temperatures? (b) What will be the actual length of the ruler be when it is used in the desert at a temperature of 40 0C? Sol. (a) If the temperature decreases, the length of the ruler also decreases through thermal contraction. Below 200C, each centimeter division is actually somewhat shorter than 1.0 cm, so the steel ruler gives readings that are too long. (b) At 400C, the increases in length of the ruler is =T = (20) (1.2 × 10–5) (400 – 200) = 0.48 × 10–2 cm  The actual length of the ruler is, '=+=20.0048 cm Ex. A second's pendulum clock has a steel wire. The clock is calibrated at 200C. How much time does the clock lose or gain in one week when the temperature is increased to 300C? (  steel  1.2  10 5 C 1 ) Sol. The time period of second's pendulum is 2 second. As the temperature increases length time period increases. Clock becomes slow and it loses the time. The change in time period is T  1 1 T  =   2  1.2  1 0 5   3 0 0  2 0 0  = 1.2 × 10–4 s 2 2  New Time period is T   T  T   2  1.2  10 4  = 2.00012 s 1 .2  10 4   T   7  2 4  36 00  = 36.28 s  Time lost in one week t    t   T 2.00 01 2  Thermal Expansion in Liquids (i) Liquids do not have linear and superficial expansion but these only have volume expansion. (ii) Since liquids are always to be heated along with a vessel which contains them so initially on heating the system (liquid + vessel), the level of liquid in vessel falls (as vessel expands more since it absorbs heat and liquid expands less) but later on, it starts rising due to faster expansion of the liquid. R P Q PQ  represents expansion of vessel QR represents the real expansion of liquid. (iii) The actual increase in the volume of the liquid = The apparent increase in the volume of liquid + the increase in the volume of the vessel. (iv) Liquids have two coefficients of volume expansion. (A) Co–efficient of apparent expansion (a) It is due to apparent (that appears to be, but in not) increase in the volume of liquid if expansion of vessel containing the liquid is not taken into account. Apparent expansion in volume (V)  Initial volume   V   a  (B) Co–efficient of real expansion (r) (a) It is due to the actual increase in volume of liquid due to heating. r  Real increase in volume (V)  Initial volume   V   (b) Also coefficient of expansion of flask  Vessel  (c) Real = Apparent + Vessel (V)Vessel V   (d) Change (apparent change) in volume in liquid relative to vessel is Vapp= V(Real – Vessel)  = V(r – 3)  = Coefficient of linear expansion of the vessel. Different level of liquid in vessel  Re al.   Vessel ( 3)   app  0 V Vapp is positive Level Level of liquid inVessel will rise on heating  Re al.  Vessel ( 3)   app  0 Vapp is negative Level of liquid in vessel will fall on heating  Re al  Vessel ( 3)   app  0 Ex. Vapp  0 Level of liquid in vessel will remain same Water at temperature t0C In figure shown, left arm of a U–tube is immersed in a hot water bath at temperature t°C, and right arm is immersed in a bath of melting ice; the height of manometric liquid in respective columns is ht and h0. Determine the coefficient of expansion of the liquid. Sol. ht Melting ice h0 The liquid is in hydrostatic equilibrium  t gh t  0 gh 0 Where, t is density of liquid in hot bath, 0 is density of liquid in cold bath. Volumes of a given mass M of liquid at temperatures t and 00C are related by Vt = V0(1+t) Since t Vt  0 V0  t  Since ht = 0 V0 0  Vt 1  t  h t  h0  0 h 0  h 0 1  t  which on solving for  , yields  = h0 t t Anomalous expansion of water Generally matter expands on heating and contracts on cooling. In case Bahaviour Anomalous Vol/mass Vol/mass min Anomalous Bahaviour of water, it expands on heating if its temperature is greater than 4°C. In the range 0°C to 4°C, water contracts on heating and expands on cooling, i.e.  is negative. This behaviour of water in the range from 0°C to 4°C is called anomalous expansion. maximum 0°C 4°C 0°C 4°C (A) Temperature (B) Temperature THERMAL PROPERTIES OF MATTER This anomalous behaviour of water causes ice to form first at the surface of a lake in cold weather. As winter approaches, the water temperature increases initially at the surface. The water there sinks because of its increased density. Consequently, the surface reaches 0°C first and the lake becomes covered with ice. Aquatic life is able to survive the cold winter as the lake bottom remains unfrozen at a temperature of about 4°C. At 4°C, density of water is maximum while its specific volume is minimum. Ex. The difference between lengths of a certain brass rod and of a steel rod is claimed to be constant at all temperatures. Is this possible ? Sol. If LB and LS are the lengths of brass and steel rods respectively at a given temperature, then the lengths of the rods when temperature is changed by  °C. LB= LB(1 + B ) and LS= LS(1 + B ) So that LB= LS(LB – LS) + (LB B – LSS)  LB S  L S B i.e., the difference in the lengths of the two rods will be independent of temperature if the lengths are in the inverse ratio of their coefficients of linear expansion. So (LB– LS) will be equal to (LB – LS) at all temperatures if , LBB – LSS = 0 [as  0] or Ex. There are two spheres of same radius and material at same temperature but one being solid while the other hollow. Which sphere will expand more if (a) they are heated to the same temperature, (b) same heat is given to them ? Sol. (a) As thermal expansion of isotropic solids is similar to true photographic enlargement, expansion of a cavity is same as if it had been a solid body of the same material i.e. V = V V V As here V,  and  are same for both solid and hollow spheres treated (cavity) ; so the expansion of both will be equal. (b) If same heat is given to the two spheres due to lesser mass, rise in temperature of hollow sphere will be more [as   Q ] and hence its expansion will be more [as V = V]. mc