Temperature & Thermal Expansion

TEMPERATURE & THERMAL EXPANSION 1. ● TEMPERATURE & TEMPERATURE SCALES Temperature Temperature may be defined as the degree of hotness or coldness of a body. Heat energy flows from a body at higher temperature to that at lower temperature until their temperatures become equal. At this stage, the bodies are said to be in thermal equilibrium. ● Thermal equilibrium Thermal equilibrium is a situation in which two objects would not exchange energy by heat or electromagnetic radiation if they were placed in thermal contact. Heat is the transfer of energy from one object to another object as a result of a difference in temperature between them. ● Zeroth law of thermodynamics If objects A and B are separately in thermal equilibrium with a third object C (say thermometer), then objects A and B are in thermal equilibrium with each other. Zeroth law of thermodynamics introduces the concept of temperature. Two objects (or systems) are said to be in thermal equilibrium if their temperatures are same. In measuring the temperature of a body, it is important that the thermometer should be in thermal equilibrium with the body whose temperature is to be measured. ● Measurement of Temperature The branch of thermodynamics which deals with the measurement of temperature is called thermometry. A thermometer is a device use to measure the temperature of a body. The substances like liquids and gases which are used in the thermometer are called thermometric substances. ● Different Scales of Temperature A thermometer can be graduated into following scales : (a) The Centigrade or Celsius scale (°C) (b) The Fahrenheit scale (ºF) (c) Kelvin scale (K) ● Comparison between Different Temperature Scales The general formula for the conversion between different temperature scales is: K  273 C F  32 X  LFP    100 100 180 UFP  LFP Where X  Reading in unknown temperature scale, LFP  Lower Fixed Point, UFP  Upper Fixed Point K C F X    Change in temperature 100 100 180 UFP  LFP ● Although the temperature of a body can be raised without limit, it can not be lowered without limit and theoretically limiting low temperature is taken to be zero of the Kelvin scale (i.e. no negative temperature on Kelvin scale is possible). ● Though when universe was created 1010 years ago, its temp. was about 1039 K which at present is about 3K. The highest laboratory temperature is about 108 K (in fusion test reactor) while lowest 1010 K (achieved in 1999 through nuclear spin cooling) Theory has established that zero Kelvin temperature can never be achieved practically. Illustrations Illustration 1. Temperature of a patient is 40ºC. find the temperature on Fahrenheit scale ? Solution: F  32 40  0  F = 104ºF  180 100 Illustration 2. At what temperature is the Fahrenheit scale reading equal to twice of Celsius ? Solution: F  32 C  0 2x  32 x  0     x = 160 180 100 180 100 Illustration 3. The lower and upper fixed points of a faulty thermometer are 5 W and 105 W. If the thermometer reads 25 W, what is the actual temperature in Celsius' scale? Solution: 25  5 C  0   C = 20ºC 100 100 Illustration 4. A thermometer with an arbitrary scale has the ice point at –20º and the steam point at 180º. When the thermometer reads 5º, a Centigrade thermometer will read (1) 7.5 ºC (2) 12.5 ºC (3) 16.5 ºC (4) –9.37 ºC Solution: C0 t  (20) (Here t = 5º)  100  0 180  (20) Illustration 5. The temperature of an iron piece is raised from 30ºC to 90ºC. What is the change in its temperature on the Fahrenheit scale and on the Kelvin scale? Solution: C = 90º – 30º = 60ºC 9 9 Temperature difference on Fahrenheit Scale F C = (60ºC) = 108ºF 5 5 Temperature difference on Kelvin Scale K = C = 60K 2. THERMAL EXPANSION When matter is heated without any change in its 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 forces are maximum in solids but minimum in gases. ● Solids can expand in one dimension (Linear expansion), two dimensions (Superficial expansion) and three dimensions (Volumetric expansion) while liquids and gases usually suffers change in volume only. ● Linear expansion :  = 0 (1 + )   = 0 ● Superficial (areal) expansion : A = A0 (1 + ) Also A0 = l 20 and A = 2 So 2 = l 20 (1 + ) = [0(1 + )]2   = 2 ● Volumetric expansion : V = V0 (1 + ) Also V = 3 and V0 = l 30 so  = 3 3  = [0 (1 + )]3  6 = 3 = 2 or  :  :  = 1 : 2 : 3 = coefficient of linear expansion = coefficient of superficial expansion = coefficient of volumetric expansion ● 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. Applications of thermal Expansion in Solids (a) Bi-metallic strip : When two strips of eq4aJlength but of different materials (different coefficient of linear expansion) are joined 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. Coefficient of expansion is more for brass than steel. (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. T 1 T 1 l   ( T  l   Fractional change in time period ) T 2 T 2 l • Due to increment in its time period, a pendulum clock becomes slow in summer and 1 will lose time. Loss of time in a time period T = T. 2 • 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). • Since coefficient of linear expansion () is very small for invar, hence pendulums are made of invar to show the correct time in all seasons. L L 1 stress  =    = So thermal stress = Y  Y = L L  strain So force on the supports F = YA Thermal strain = (d) Expansion of cavity : Thermal expansion of an isotropic object may be imagined as a photographic enlargement. (e) • • • Some other applications When rails are laid down on the ground, space is left between the ends of two rails The transmission cable are not tightly fixed to the poles Test tubes, beakers and cubicles are made up of pyrex-glass or silica because they have very low value of coefficient of linear expansion • • ● The iron rim to be put on a cart wheel is always of slightly smaller diameter than that of wheel A glass stopper jammed in the neck of a glass bottle can be taken out by warming the neck of the bottle. Thermal Expansion in Liquids • Liquids do not have linear and superficial expansion but these only have volumetric expansion. • 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. QP represents expansion of vessel QR represents the real expansion of liquid. • 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. • Liquids have two coefficients of volume expansion. (i) Co-efficient of apparent expansion (a) It is due to apparent (that appears to be, but not in real) increase in the volume of liquid if expansion of vessel containing the liquid is not taken into account. Apparent expansion in volume ( V)  a = Initial volume   V   (ii) Co-efficient of real expansion (r) It is due to the actual increase in volume of liquid due to heating. Re alincrease in volume ( V)  r = Initial volume   V   • Also coefficient of expansion of flask Vessel = • • (V) Vessel V   Real = Apparent + Vessel 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 GOLDEN KEY POINTS ● ● Actually thermal expansion is always 3-D expansion. When other two dimensions of object are negligible with respect to one, then observations are significant only in one dimension and it is known as linear expansion. Relation between ,  and       (i) For isotropic solids:  :  :  = 1: 2 : 3 or 1 2 3 (ii) For non-isotropic solids  = 1 + 2 and  = 1 + 2 + 3. Here 1 , 2 and 3 are coefficient of linear expansion in X, Y and Z direction. Illustrations Illustration 6. A rectangular plate has a circular cavity as shown in the figure. If we increase its temperature then which dimension will increase in following figure. Solution: Distance between any two point on an object increases with increase in temperature. So, all dimension a, b, c and d will increase Illustration 7. A small ring having small gap is shown in figure on heating what will happen to the size of gap. Solution: Gap will also increase. The reason is same as in above example. Illustration 8. What is the percentage change in length of 1 m iron rod if its temperature changes by 100ºC.  for iron is 2 × 10–5/ºC. Solution: Percentage change in length due to temperature change l % = × 100 =  × 100 = 2 × 10–5 × 100 × 100 = 0.2% l Illustration 9. A concrete slab has a length of 10 m on a winter night when the temperature is 0ºC. Find the length of the slab on a summer day when the temperature is 35ºC. The coefficient of linear expansion of concrete is 1.0 × 10–5 /ºC. Solution: t = 10(1 + 1 × 10–5 × 35) = 10.0035 m Illustration 10. A glass vessel of volume 100 cm3 is filled with mercury and is heated from 25°C to 75°C. What volume of mercury will overflow? Coefficient of linear expansion of glass = 1.8 × 10–6/ºC and coefficient of volume expansion of mercury is 1.8 × 10–4/ºC. Solution: V = v0(L – L) T = 100 × {1.8 × 10–4 – 3 × 1.8 × 10–6} × 50 V = 0.87 cm3 Illustration 11. 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 they are heated to the same temperature, Solution: 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  As here V,  and  are same for both solid and hollow spheres treated (cavity) ; so the expansion of both will be equal. Illustration 12. A steel wire of cross-sectional area 0.5 mm2 is held between two fixed supports. If the wire is just taut at 20ºC, determine the tension when the temperature falls to OºC. Coefficient of linear expansion of steel is 1.2 × 10–5/ºC and its Young's modulus is 2.0 × 1011 N/m2. Solution: Here final length is less than the original length so that strain is tensile and tensile force is given by F = AY = 0.5 × 10–6 × 2 × 1011 × 1.2 × 10–5 × 20 = 24 N 1. BEGINNER’S BOX - 1 Write down the following temperatures !n the increasing order 50ºF, 50ºC and 50 K. 2. The figure shows three temperature scales with the freezing and boiling points of water indicated. 3. (a) Rank the size of a degree on these scales, greatest first. (b) Rank the following temperatures, highest first : 50ºX, 50ºW and 50ºY. What is the temperature at which we get the same reading on both the Centigrade and Fahrenheit scales? 4. A thin copper wire of length L increases in length by 1% when heated from temperature T 1 to T2. What is the percentage change in area when a thin copper plate having dimensions 2L × L is heated from T1 to T2 ? (A) 1% (B) 2% (C) 3% (D) 4% 5. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C ? Coefficient of linear expansion of copper = 1.70 × 10–5 ºC–1 6. If a bimetallic strip is heated, it will (A) bend towards the metal with lower thermal expansion coefficient. (B) bend towards the metal with higher thermal expansion coefficient. (C) twist itself into helix. (D) have no bending.