This is the teaching module for Elasticity, Calorimetry & Thermal Expansion . This module has to be followed in the class. VK Bansal
TEACHING NOTES ELASTICITY, CALORIMETRY & THERMAL EXPANSION (Complete in 3 Lectures)
ELASTICITY The shape of a solid can be altered only by the application of a suitable system of forces. ELASTICITY: If the modified shape is retained by the solid only so long as the forces act, and the original shape is regained when they cease to act, the solid is said to possess elasticity ; PLASTICITY: if the modified shape is retained after the forces cease to act, the solid is said to be plastic. >Fluids, including liquids and gases, possess no definite shape, and therefore cannot possess elasticity. > The volume of matter, whether this matter be solid or fluid, can be altered by the application ofsuitable forces, and in most cases the original volume is regained when the forces cease to act ; hence, all matter possesses volume elasticity. >Any alteration produced in the shape or volume of matter is called a strain. Cause of elasticity We know that in a solid, each atom or molecule is surrounded by neighboring atoms or molecules. These are bonded together by interatomic or intermolecular forces and stay in a stable equilibrium position. When a solid is deformed, the atoms or molecules are displaced from their equilibrium positions causing a change in the interatomic (or intermolecular) distances. When the deforming force is removed, the interatomic forces tend to drive them back to their original positions. Thus the body regains its original shape and size. The restoring mechanism can be visualised by taking a model of spring-ball system shown in the Fig.. Here the balls represent atoms and springs represent interatomic forces.
Fig. Spring-ball model for the illustration of elastic behaviour of solids. If you try to displace any ball from its equilibrium position, the spring system tries to restore the ball back to its original position. Thus elastic behaviour of solids can be explained in terms of microscopic nature of the solid.When a body is subjected to a deforming force, a restoring force is developed in the body. This restoring force is equal in magnitude but opposite in direction to the applied force.
Page # [1]
Stress : Restoring force per unit area set up inside the body is called stress and is measured by the magnitude of force acting on unit area of the body in equilibrium. If F is the force applied and A is the area of cross section of the body, Magnitude of the stress = F/A. The SI unit of stress is N m–2 or pascal (Pa) and its dimensional formula is [ ML–1T–2 ]. Stress (Tensile/Compressive ) Tensile
F
Compressive
F
F
F
F
=
F as F A A
F
Strain (longitudinal) Suppose we stretch a wire by applying tensile forces of magnitude F to each end. The length of the wire increases from L to L + L. The fractional length change is called the strain ; it is a dimensionless measure of the degree of deformation. strain =
L L
Hooke’s law for tensile and compressive forces Suppose we had wires of the same composition and length but different thicknesses. It would require larger tensile forces to stretch the thicker wire the same amount as the thinner one. We conclude that the tensile force required is proportional to the cross-sectional area of the wire (F A). Thus, the same applied force per unit area produces the same deformation on wires of the same length and composition. Hooke’s Law stress strain F L =Y A L equation still says that the length change (L) is proportional to the magnitude of the deforming forces (F). Stress and strain account for the effects of length and cross-sectional area ; the proportionality constant Y depends only on the inherent stiffness of the material from which the object is composed ; it is independent of the length and cross-sectional area. F L L =Y k =Y A. Y is called the elastic modulus or Young’ss A L L modulus, Y has the same units as those of stress (Pa or N/m2) since strain is dimensionless. Young’s modulus can be through of as the inherent stiffness of a material ; it measures the resistance of the material to elongation or compression. Material that is flexible and stretches easily (for example, rubber) has a low Young’s modulus. A stiff material (such as steel) has a high Young’s modulus ; it takes a larger stress to produce the same strain. Hooke’s law holds up to a maximum stress called the proportional limit. For many materials, Young’s modulus has the same value for tension and compression. Some composite materials, such as bone and concrete, have significantly different Young’s moduli for tension and compression. The different properties of these two substances lead to different values of Young’s modulus for tension and compression. Important Notes : 1. Y is independent of size, shape of body depends only on nature and material 2. Ysolid > Yliq. > Ygas Yrigid =
Comparing equation F = kL and
3.
genrally as Tempertureincreases Y
Page # [2]
Ex. [Sol.
A light wire of length 4m is suspended to the ceiling by one of its ends. If its crossectional area is19.6 mm2, what is its extension under a load of 10kg. Young’s modulus of steel = 2 × 1011 Pa. Given ; original length L = 4m ; force F = 10 × 9.8 = 98 N ; and Y = 2 × 1011 Nm–2 ; l = ? Using the relation, longitudin al stress Young’s modulus (Y) = Iongitudin al strain
We have F/A Y= l/L
FL YA
98 4 l = 2 1011 19 . 6 10 6 = 1 × 10–4 m
= 0.1 mm Ex.
l=
]
Two vertical rods of equal lengths, one of steel and the other of copper, are suspended from the ceiling, at a distance l apart and are connected rigidly to a rigid horizontal light bar at their lower ends.
Steel
Copper
F
[Sol.
If AS and AC be their respective cross sectional areas, and YS and YC, their respective Young’s modulii of elasticities, find where should a vertical force F be applied to the horizontal bar, in order that the bar remains horizontal. (Fig.) Let the force F be applied at a distance x from the steel bar, measured along the horizontal bar. Let FS and FC be the loads on steel and copper rods respectively, so FS + FC = F ... (i) Since the rigid horizontal bar remains horizontal so, the extensions produced in the two rods and hence strains remains same. i.e.,
FS FC = AS YS A C YC
... (ii)
FA S YS Solving (i) and (ii) FS = A Y A Y S S C C
and
FA C YC FC = A Y A Y S S C C
Now, taking moments about the steel bar. FC l = Fx
or
x = l / 1
x= AS AC
FC l F
YS YC
A C YCl A Y A Y S S C C
]
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Elastic potential energy (where F is applied in equilibrium) AY x F L
dw = Fdx =
AY x dx L AY L
w=
F F
F
l+x
l
x dx 0
AY l 2 U= 2L
U=
F
=
1 yl l (AL) 2 L L
1 (stress) (strain) (volume) 2
Optional Ex. A wire having a length l = 2m, and cross sectional area A = 5mm2 is suspended at one of its ends from a ceiling. What will be its strain energy due to its own weight, if the density and Young’s modulus of the material of the wire be d = 9g/cm3 and Y = 1.5 × 1011 Nm–2? Sol. Consider an elemental length of the wire of length dx, at a distance x from the lower end. Clearly, this length is acted upon by the external force equal to the weight of the portion of wire below it = xAdg. In equilibrium, the restoring force f = xAdg. f = xdg. A Now, elastic potential energy stored in the elemental length will be
stress =
dU =
1 × stress × strain × volume 2
=
1 (stress) 2 × × volume 2 Y
=
1 1 d 2g 2 A 2 ( xdg ) 2 × × Adx = x dx 2 2 Y Y
F Area = A
Total elastic potential energy U = dU l
1 d 2g 2 A = 2 Y x2 dx 0
1 2 2 Al 3 dg 6 Y Substituting the values,
=
f dx xAdg
x
F
(9 103 ) (9 · 8) 2 5 10 6 23 1 U= × 1.5 1011 6 = 3.46 × 10–7 J ]
Page # [4]
STRESS-STRAIN CURVE The relation between the stress and the strain for a given material under tensile stress can be found experimentally. The applied force is gradually increased in steps and the change in length is noted. A graph is plotted between the stress and the strain produced. The stress-strain curves vary from material to material. These curves help us to understand how a given material deforms with increasing loads. From the graph, we can see that in the region between O to A, the curve is linear. In this region, Hooke’s law is obeyed. The body regains its original dimensions when the applied force is removed. In this region, the solid behaves as an elastic body.
Fig. Stress-strain curve for a metal. Beyond hooke’s law In the region from A to B, stress and strain are not proportional. Nevertheless, the body still returns to its original dimension when the load is removed. The point B in the curve is known as yield point (also known as elastic limit) and the corresponding stress is known as yield strength (Sy) of the material. If the tensile or compressive stress exceeds the proportional limit, the strain is no longer proportional to the stress (figure). The solid still returns to its original length when the stress is removed as long as the stress does not exceed the elastic limit. If the stress exceeds the elastic limit, the material is permanently deformed. For still larger stresses, the solid factures when the stress reaches the breaking point. The maximum stress that can be withstood without breaking is called the ultimate strength. The ultimate strength can be different for compression and tension ; then we refer to the compressive strength or the tensile strength of the material. A ductile material continues to stretch beyond its ultimate tensile strength without breaking ; the stress then decreases from the ultimate strength (fig. (a) ). Examples of ductile solids are relatively soft metals, such as gold, silver, copper, and lead. These metals can be pulled like taffy, becoming thinner and thinner until finally reaching the breaking point. Ductile Tensile stress
Ultimate Elastic strength limit • • • Breaking point • Proportional limit
Brittle Tensile stress
Ultimate strength and breaking point • •Elastic limit • Proportional limit
Tensile strain Tensile strain
(b)
(a)
{Home Work - HCV : Chapter -14 Ex 1 -11 }
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DIFFERENT TYPES OF STRESS
Fig. (a) Cylinder subjected to tensile stress stretches it by an amount L. (b) A cylinder subjected to shearing (tangential) stress deforms by an angle . (c) A book subjected to a shearing stress (d) A solid sphere subjected to a uniform hydraulic stress shrinks in volume by an amount V. There are three ways in which a solid may change its dimensions when an external force acts on it. These are shown in Fig. In Fig.(a), a cylinder is stretched by two equal forces applied normal to its crosssectional area. The restoring force per unit area in this case is called tensile stress. If the cylinder is compressed under the action of applied forces, the restoring force per unit area is known as compressive stress. Tensile or compressive stress can also be termed as longitudinal stress. In both the cases, there is a change in the length of the cylinder. The change in the length L to the original length L of the L body (cylinder in this case) is known as longitudinal strain. Longitudinal strain . L Shearing stress. However, if two equal and opposite deforming forces are applied parallel to the cross-sectional area of the cylinder, as shown in Fig. (b), there is relative displacement between the opposite faces of the cylinder. The restoring force per unit area developed due to the applied tangential force is known as tangential or shearing stress. As a result of applied tangential force, there is a relative displacement x between opposite faces of the cylinder as shown in the Fig. (b). The strain so produced is known as shearing strain and it is defined as the ratio of relative displacement of the faces x to the length of the cylinder L. x Shearing strain = tan L where is the angular displacement of the cylinder from the vertical (original position of the cylinder). Usually is very small, tan is nearly equal to angle , (if = 10°, for example, there is only 1% difference between and tan ). It can also be visualised, when a book is pressed with the hand and pushed horizontally, as shown in Fig. (c). Thus, shearing strain = tan Bulk Modulus In Fig. (d), a solid sphere placed in the fluid under high pressure is compressed uniformly on all sides. The force applied by the fluid acts in perpendicular direction at each point of the surface and the body is said to be under hydraulic compression. This leads to decrease in its volume without any change of its geometrical shape. The body develops internal restoring forces that are equal and opposite to the forces applied by the fluid (the body restores its original shape and size when taken out from the fluid). The internal restoring force per unit area in this case is equal to the hydraulic pressure (applied force per unit area). The strain produced by a hydraulic pressure is called volume strain and is defined as the ratio of change in volume (V) to the original volume (V). v Volume strain v
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We have seen that when a body is submerged in a fluid, it undergoes a hydraulic stress (equal in magnitude to the hydraulic pressure). This leads to the decrease in the volume of the body thus producing a strain called volume strain. Volume Deformation Since the fluid presses inward on all sides of the object (figure), the solid is compressed-its volume is reduced. The fluid pressure P is the force per unit surface area ; it can be through of as the volume stress on the solid object. Pressure has the same units as the other kinds of stress: N/m2 or Pa.
F=PA3
F=PA2 F=PA1 F=PA3
F=PA1 F=PA2
Fig. Forces on an object when submerged in a fluid F =P A The resulting deformation of the object is characterized by the volume strain, which is the fractional change in volume :
volume stress = pressure =
change in volume V volume strain = original volume = V Unless the stress is too large, the stress and strain are proportional within a constant of proportionally called the bulk modulus B. A substance with a large bulk modulus is more difficult to compress than a substance with a small bulk modulus. For solids and liquids, the volume strain due to atmospheric pressure is, for most purposes, negligibly small (5 × 10–5 for water). Hooke’s law for volume deformation
P = – B
V V
V allows the V bulk modulus to be positive-an increase in the volume stress causes a decrease in volume, so V is negative. Table lists bulk moduli for various substances. The reciprocal of the bulk modulus is called compressibility and is denoted by k. It is defined as the fractional change in volume per unit increase in pressure. k = (1/B) = – (1/p) × (V/V) Bulk modulus (all three states of matter)
where V is the volume at atmospheric pressure. The negative sign. equation P = – B
B=
volume stress strain
P P B = V = – V V V
C=
1 1 V =– B V P
Page # [7]
The bulk moduli of liquids are generally not much less than those of solids, since the atoms in liquids are nearly as close together as those in solids. Gases are much easier to compress than solids or liquids, so their bulk moduli are much smaller. The bulk moduli of a few common materials are given in Table
Table : Bulk moduli (B) of some common Materials Poisson’s ratio : When an elongation is produced by longitudinal stresses, a change is produced in the lateral dimensions of the strained substance. Thus, when a wire is stretched, its diameter diminishes ; and when the longitudinal strain is small, the lateral strain is proportional to it. The ratio of the lateral strain to the longitudinal strain is called Poisson’s ratio.
F1
l3 l2
F1
l1
F A Y = l 1 l1 l1 l2 l = 3 =– l l2 l3 1 Results y B = 3 (1 2)
Page # [8]
Optional: Relation between bulk modulus & Young’s modulus of a bar is in fluid z l3
l2
F1
x y
l1 using superposition (i)
l1 P = – y l1 x
l1 = l
(ii)
P l2 =– y l
by symmetry
l1 = – l2 = + P y l1 y l
P l 2 =– (1 – 2) y l
l1 l1 z
P l3 =– (1 – 2) y l for small change in volume
(iii)
= +s
P y
P l = – (1 – 2) Y x y z l
P l3 =– y l
V l1 l2 l3 + + = V l l l
l1 P = + y l1 z
P V = – 3 (1 – 2) V y
Ex.
A uniform bar of length L and cross sectional area A is subjected to a tensile load F. If Y be the Young’s modulus of the material of the bar and be its poisson’s ratio, then determine the volumertic strain.
[Sol.
Longitudinal stress =
F . A
F = 1 (say) AY Now, by definition of Poisson’s ratio,
Longitudinal strain =
... (i)
lateral strain r / r = longitudin al strain = L / L F [From eqn. (i) ] AY Since Volumetric strain = Strain in length + Twice strain in radius.
or
r / r = – L / L
Volumetric strain = =
–
L 2r + L r
F F 2 AY AY
{Home Work - HCV : Chapter -14
=
F (1 – 2). AY
]
Ex 12 - 15, Question & Short Answer 1 – 6 Obj. - I 1 – 10 Obj. - II 1 } Page # [9]
CALORIMETRY
Ex.
Ans.
Ex.
Sol.
A system is said to be isolated if no exchange or transfer of heat occurs between the system and its surroundings. When different parts of an isolated system are at different temperature, a quantity of heat transfers from the part at higher temperature to the part at lower temperature. The heat lost by the part at higher temperature is equal to the heat gained by the part at lower temperature. Calorimetry means measurement of heat. When a body at higher temperature is brought in contact with another body at lower temperature, the heat lost by the hot body is equal to the heat gained by the colder body, provided no heat is allowed to escape to the surroundings. A device in which heat measurement can be made is called a calorimeter. It consists a metallic vessel and stirrer of the same material like copper or aluminium. The vessel is kept inside a wooden jacket which contains heat insulating materials like glass wool etc. The outer jacket acts as a heat shield and reduces the heat loss from the inner vessel. There is an opening in the outer jacket through which a mercury thermometer can be inserted into the calorimeter. The following example provides a method by which the specific heat capacity of a given solid can be determinated by using the principle, heat gained is equal to the heat lost. Water equivalent of a calorimeter means amount of heat required for the equivalent amount of water to raise temperature by 1°C. A sphere of aluminium of 0.047 kg placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100 °C. It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg of water at 20 °C. The temperature of water rises and attains a steady state at 23 °C. Calculate the specific heat capacity of aluminium. In solving this example we shall use the fact that at a steady state, heat given by an aluminium sphere ill be equal to the heat absorbed by the water and calorimeter. Mass of aluminium sphere (m1) = 0.047 kg Initial temp. of aluminium sphere = 100 °C Final temp. = 23 °C Change in temp ( T ) = (100 °C – 23 °C) = 77 °C Let specific heat capacity of aluminium be sAl. The amount of heat lost by the aluminium sphere = m1sAlT = 0.047 kg × sAl × 77°C Mass of water (m2) 0.25 kg Mass of calorimeter (m3) = 0.14 kg Initial temp. of water and calorimeter = 20°C Find temp. of the mixture = 23°C Change in temp. (T2) = 23°C – 20°C = 3°C Specific heat capacity of water (sw) = 4.18 × 103 J kg–1 K–1 Specific heat capacity of copper calorimeter = 0.386 × 103 J kg–1 K–1 The amount of heat gained by water and calorimeter = m2 sw T2 + m3scu T2 = (m2sw + m3scu) (T2) = (0.25 kg × 4.18 × 103 J kg–1 K–1 + 0.14 kg × 0.386 × 103 J kg–1 K–1 ) (23°C – 20°C) In the steady state heat lost by the aluminium sphere = heat gained by water + heat gained by calorimeter. So, 0.047 kg × sN × 77°C = (0.25 kg × 4.18 × 103 J kg–1 K–1 + 0.14 kg × 0.386 × 103 J kg–1 K–1) (3°C) sAl = 0.911 kJ kg–1 K–1 A 5g piece of ice at – 20°C is put into 10g of water at 30°C. Assuming that heat is exchanged only between the ice and the water, find the final the final temperature of the mixture. Specific heat capacity of ice = 2100 J/kg-°C specific heat capacity of water = 4200 J/kg-°C and latent heat of fusion of ice = 3.36 × 105 J/kg. The heat given by the water when it cools down from 30°C to 0°C is (0.01 kg) (4200 J/k-°C) (30°C) = 1260 J. The heat required to bring the ice to 0°C is (0.005 kg) (2100 J/kg-°C) (20°C) = 210 J. The heat required to melt 5g of ice is (0.005 kg) (3.36 × 105 J/kg) = 1680. We see that whole of the ice cannot be melted as the required amount of heat is not proved by the water. Also, the heat is enough to bring the ice to 0°C. Thus the final temperature of the mixture is 0°C with some of the ice melted. Page # [10]
THERMAL EXPANSION Thermal Expansion When matter is heated without change in 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 say from T1 to T2 the amplitude of vibration and hence energy of atoms increases from E1 to E2 and hence the average distance between the from r1 to r2. + 0 E2 E E1 E0
T2 r2 r1 T1 r0 T0 r
Due to this increase in distance between atoms, the matter as a whole expands. Had the potential energy curve been symmetrical, no thermal expansion would have taken place in spite of heating. Normal Solids To varying extents, most materials expand when heated and contract when cooled. The increase in any one dimension of a solid is called linear expansion, linear in the sense that the expansion occurs along a line. A rod whose length is L0 when the temperature is T0 when the temperature increases to T0 + T, the length becomes L0 + L, where T and L are the magnitudes of the changes in temperature and length, respectively. Conversely, when the temperature decreases to T0 – T, the length decreases to L0 – L. For small temperature changes, experiments show that the change in length is directly proportional to the change in temperature (L T). In addition, the change in length is proportional to the initial length of the rod, L0 L
Equation L = L0T expresses the fact that L is proportional to both L0 and T(L L0T) by using a proportionality constant , which is called the coefficient of linear expansion. 1 Common unit for the coefficient of linear expansion : = (C°)–1 C Ex. [Sol.
A brass scale correctly calibrated at 15°C is employed to measure a length at a temperature of 35°C. If the scale gives a reading of 75cm, find the true length. (Linear expansively of brass = 2.0 × 10–5C°)–1) Let the distance between two fixed divisions on the scale at 15°C be L1 and that at 35°C be L2. Clearly, (L2 – L1) = L1 (35 – 15) or L2 = L1 (1 + 20 × 2.0 × 10–5) = L1 (1.0004) i.e., at 35°C, an actual length of L2 will be read as L1 (note), due to the increased separation of the divisions of the scale. In other words, the observed length will be less than the actual length. Given : L1 = 75 cm L2 = 75 (1.0004) cm = 75.03 cm ] Page # [11]
Ex.
[Sol.
Estimate the time lost or gained by a pendulum clock at the end of a week when the atmospheric temperature rises to 40°C. The clock is known to give correct time at 15°C and the pendulum is of steel. (Linear expansively of steel is 12 × 10–6 / °C). Rise in temperature = 40 – 15 = 25°C Time lost per second =
1 2
1 × (12 × 10–6 / °C) × (25°C) 2 = 150 × 10–6 s/s Therefore, time lost per week (i.e., 7 × 86400 s) = 150 × 10–6 s/s × 7 × 86400 s = 90.72 s ]
=
Ex.
Sol.
A glass rod when measured with a zinc scale, both being at 30°C, appears to be of length 100cm. If the scale shows correct reading at 0°C, determine the true length of the glass rod at (a) 30°C and (b) 0°C. (‘’ for glass = 8 × 10–6 /°C and for zinc 26 × 10–6 / °C) At 30°C, although the reading shown by the zinc scale corresponding to the length of the glass rod is 100cm, but the actual length would be more than 100cm, the reason being the increased separation between the markings, owing to a rise in temperature (from 0°C to 30°C). Now, an actual (original at 0°C) length of 100cm on the zinc scale (or more precisely, two markings or divisions on the scale, separated by a distance of 100cm) would, at a temperature of 30°C, correspond to a length given by l = 100 (1 + 26 × 10–6 × 30) cm = 100.078 cm The true length of the glass rod at 30°C is 100.078 cm. Now, at 0°C, the length of glass rod would be lesser than that at 30°C,
lt Using lt = l0 (1 + t), l0 = 1 t
The length of the rod at 0°C, will be l0 =
100.078 cm = 100.054 cm ] (1 8 10 6 30)
The Bimetallic Strip A bimetallic strip is made from two thin strips of metal that have different coefficients of linear expansion, as figure shows. Brass Steel
(a)
(b) Heated (c) Cooled
(a) A bimetallic strip and how it behaves when (b) heated and (c) cooled
Page # [12]
Often brass [ = 19 × 10–6 (C°)–1] and steel [ = 12 × 10–6 (C°)–1] are selected. The two pieces are welded or riveted together. When the bimetallic strip is heated, the brass, having the larger value of , expands more than the steel. Since the two metals are bonded together, the bimetallic strip bends into an arc as in part b, with the longer brass piece having a larger radius than the steel piece. When the strip is cooled, the bimetallic strip bends in the opposite direction, as in part c. Mathematical Solution l
d R = L0 (1 + 1) 2
d d
d R = L0 (1 + 2) 2 R
d 1 1 2 = d 1 2 R 2 R
(R–d/2)
(R+d/2)
Find R {Home Work - HCV : Chapter - 25 } The expansion in length is called linear expansion. The expansion in area is called area expansion. The expansion in volume is called volume expansion
Ex.
=
=
=
Show that the coefficient of area expansion, (A/A)/T. of a rectangular sheet of the solid is twice its linear expansively, 1.
Ans.
Consider a rectangular sheet of the solid material of length a and breadth b (Fig.). When the temperature increases by T, a increases by a = a 1T and b increases by b = b T. From figure the increase in area Page # [13]
A = A1 + A2 + A3 A = a b + b a + (a) (b) = a 1b T + b 1 aT + (1)2 ab (T)2 = 1ab T (2 + 1 T) = 1 A T (2 + 1 T) –5 –1 Since 1 ; 10 K , the product 1 T for fractional temperature is small in comparison with 2 and may be neglected. Hence, A 1 = 21 A T
Volume Thermal Expansion The volume of a normal material increases as the temperature increases. Most solids and liquids behave in this fashion. By analogy with linear thermal expansion, the change in volume V is proportional to the change in temperature T and to the initial volume V0, provided the change in temperature is not too large. These two proportionalities can be converted into equation V = V0T with the aid of a proportionality constant , known as the coefficient of volume expansion. The algebraic form of this equation is similar to that for linear expansion, L = L0T. Volume thermal expansion The volume V0 of an object change by an amount V when its temperature changes by an amount T. V = V0T where is the coefficient of volume expansion. Common Unit for the coefficient of volume Expansion : (C°)–1 The unit for , like that for , is (C°)–1. Values for depend on the nature of the material. The values of for liquids are substantially larger than those for solids, because liquids typically expand more than solids, given the same initial volumes and temperature expansion is three times greater than the coefficient of linear expansion : = 3. If a cavity exists within a solid object, the volume of the cavity increases when the object expands, just as if the cavity were filled with the surrounding material. The expansion of the cavity is analogous to the expansion of a hole in a sheet of material. Accordingly, the change in volume of a cavity can be found using the relation V = V0T, where is the coefficient of volume expansion of the material that surrounds the cavity. The expansion of holes An interesting example of linear expansion occurs when there is a hole in a piece of solid material. We know that the material itself expands when heated. But what about the hole? Does it expand, contract, or remain the same? Following example provides some insight into the answer of this question Example Do holes expand or contract when the temperature increases? Figure (a) shows eight square tiles that are arranged to form a square pattern with a hole in the centre. If the tiles are heated, what happens to the size of the hole? Hole
Expanded hole
9th tile (heated)
(a) unheated
(b) Heated
(c)
Page # [14]
Reasoning and solution We can analyze this problem by disassembling the pattern into separate tiles, heating, it is evident from figure (b) that the heated pattern expands and so does the hole in the centre. In fact, if we had a ninth tile that was identical to and also heated like the others, it would fit exactly into the centre hole, as figure (c) indicates. Thus, not only does the hole in the pattern expand, but it expands exactly as much as one of the tiles. Since the ninth tile is made of the same material as the others, we see that the hole expands just as if it were made of the material of the surrounding tiles. The thermal expansion of the hole and the surrounding material is analogous to a photographic enlargement ; in both situations everything is enlarged, including holes. Thus, it follows that a hole in a piece of solid material expands when heated and contracts when cooled, just as if it were filled with the material that surrounds it. If the hole is circular, the equation L = L0T can be used to find the change in any linear dimension of the hole, such as its radius or diameter. Example illustrates this type of linear expansion. Asking question Figure shows a cross-sectional view of three cylinders, A, B and C. Each is made from a different material ; one is lead, one is brass, and one is steel. All three have the same temperature, and barely fit inside each other. As the cylinders are heated to the same, but higher, temperature, cylinder C falls off, While cylinder A becomes tightly wedged to cylinder B. Given, lead has the greatest coefficient of linear expansion, followed by brass, and then by steel.Which cylinder is made from which material? Brass
Lead
A
A
Steel
Steel
B
B C
Lead
(a)
Brass
C
(b)
Reasoning and solution We need to consider how the outer and inner diameters of each cylinder change as the temperature is raised. With respect to the inner diameter, we will be guided by the fact that a hole expands as if it were filled with the surrounding material. These data indicate that the outer and inner diameters of the lead cylinder change the most, while those of the steel cylinder change the least. Since the steel cylinder expands the least, it cannot be the outer one, for if it were, the greater expansion of the middle cylinder would prevent the steel cylinder from falling off. The steel cylinder also cannot be the inner one, because then the greater expansion of the middle cylinder would allow the steel cylinder to fall out, contrary to what is observed. The only place left for the steel cylinder is in the middle, which leads to the two possibilities in figure. In part a, lead is on the outside and will fall off as the temperature is raised, since lead expands more than steel. On the other hand, the inner brass cylinder expands more than the steel that surrounds it and becomes tightly wedged, as observed. Thus, one possibility is A = brass, B = steel, and C = lead. In part b of the drawing, brass is on the outside. As the temperature is raised, brass expands more than steel, so the outer cylinder will again fall off. The inner lead cylinder has the greatest expansion and will be wedged against the middle steel cylinder. A second possible answer, then is A = lead, B = steel, and C = brass.
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Thermal Stress Ex. A brass rod of length 1m is fixed to a vertical wall, at one end, with the other end free to expand. When the temperature of the rod is increased by 120°C, the length increases by 2cm. What is the strain? Sol. After the rod expands, to the new length there are no elastic forces developed internally in it. So, strain = 0. F
L D
top view
D – D
L F
Note : A change in shape/size i.e., dimensions need not necessarily imply a strain. For example, if a body is heated to expand, its volume change, as it acquires a new size, due to expansion. However, the strain remains zero. Unless and until, internal elastic forces operate, to bring the body to the original state, no strain exists. When a body is heated, the total energy of molecule increase, owing to an increase in the kinetic energy of the molecules. This results in a shift (increment) of the “equilibrium distance” of molecules and the body acquires a new shape and size, in the expanded form, whereby the molecules are in “zero force” state. Hence, there is no strain. However, if the body is resistricted to expand, during the process of heating, then the molecules become “strained”, and even if there is no apparent change in dimensions of the body, there is strain. In such cases, strain is measured as the ratio. In dimension that would have occured, and the change in dimension that would have occured, had the body been free to expand or contract, to the original dimension. When a metal rod is heated or cooled it tends to expand or contract. If it is left free to expand or contract, no temperature stresses will be induced. However, if the rod be restricted to change its length, then temperature stresses are generated within it. Stress induced due to temperature change can be understood as follows: B
A L
l
F
F
Consider a uniform rod AB fixed rigidly between two supports. (fig.) If L be its length, the coefficient of linear expansion, then a change in temperature of , would tend to bring a change in its length by l = L. Had the rod been free (say one of its ends) its length would have changed by l. Now, let a force be gradually applied so as to restore the natural length. Since the rod, tends to remain in the new state, due to a change in temperature, so when a force F is applied, thermal stress is induced. In equilibrium. l F = (L l ) Y A Neglecting l in comparison to L,
[ stress = strain × Y]
lA Y = AY L Now, if the two ends remain fixed, then this external force is provided from the support.
F=
Clearly strain =
= L
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Ex.
The stress on a steel beam A steel beam is used in the roadbed of a bridge. The beam is mounted between two concrete supports when the temperature is 22°C, with no room provided for thermal expansion.
Beam
Concrete support
Sol.
Concrete support
What compressional stress must the concrete supports apply to each end of the beam, if they are to keep the beam from expanding when the temperature rises to 42°C? Reasoning Recall from that the stress( force per unit cross-sectional area or F/A) required to change the length L0 of an object by an amount L is L Stress = Y L 0
where Y is Young’s modulus. If the steel beam were free to expand because of the change in temperature, the length would change by L = L0T. Because the concrete supports do not permit any expansion, they must supply a stress to compress the beam by an amount L. Thus, L L 0 T Stress = Y L = Y L = YT 0 0
Young’s modulus and the coefficient of linear expansion for steel are Y = 2.0 × 1011 N/m2 and = 12 × 10–6 (C°)–1, respectively. The change in temperature from 22 to 42°C is T = 20°C. The thermal stress is Stress = YT = (2.0 × 1011 N/m2) [12 × 10–6(C°)–1] (20C°) = 4.8 × 107 N/m2] Ex.
Sol.
A rod of length 2m is at a temperature of 20°C. Find the free expansion of the rod, if the temperature is increased to 50°C. Find the temperature stresses produced when the rod is (i) fully prevented to expand, (ii) permitted to expand by 0.4 mm. Y = 2 × 1011 Nm–2 ; = 15 × 10–6 per °C. Free expansion of the rod = = (15 × 10–6 /°C) × (2m) × (50° – 20°)C = 9 × 10–4 m = 0.9 mm (i) If the expansion is fully prevented 9 10 4 4.5 × 10–4 2 temperature stress = strain × Y = 4.5 × 10–4 × 2 × 1011 = 9 × 107 Nm–2 (ii) If 0.4 mm expansion is allowed, then length restricted to expand = 0.9 – 0.4 = 0.5 mm
then strain =
Ex.
5 10 4 = 2.5 × 10–4 2 Temperature stress = strain × Y = 2.5 × 10–4 × 2 × 1011 = 5 × 107 Nm–2
Strain =
Two rods made of different materials are placed between massive walls. The cross section of the rods are A1 and A2, their lengths L1 and L2 coefficients of linear expansion 1 and 2 and the modulii of elasticity of their materials Y1 and Y2 respectively. If the rods are heated by t°C, find the force F with which the rods act on each other. Page # [17]
Sol.
Let the first rod expand slightly (say by length l ) and the second rod get compressed by the same amount (since net elongation / compression of the rods is zero.) Natural increase in length of the first rod (after being heated) when free to expand would have been 1L1t. The expansion allowed is just l (where l < 1L1t). Amount of elongation restricted = 1L1t – l
Strain =
elongation restricted 1L1t l = original length L1 (1 1t )
Since 1t << 1 1 + 1t ~ 1 1L1t l L1
Strain =
1 L1t l Y . Stress = strain × Y = L1 1
or
F = stress × A =
Similarly, F =
(1 L1t l ) Y1A1 L1
... (i)
( 2 L 2 t l ) y2t2 L2
or
FL1 FL 2 l = 1L1t – Y A = Y L – 2L2t 1 1 1 2
or
F=
(1L1 2 L 2 ) t L1 L2 Y1A1 Y1 A 2
]
FLUID EXPANSION Expansion of liquids Like solids, liquids also, in general, expand on heating ; however, their expansion is much large compared to solids for the same temperature rise. A noteworthy point to be taken into account during the expansion of liquid is that they are always contained in a vessel or a container and hence the expansion of the vessel also comes into picture. Further, linear or superficial expansion in case of a liquid does not carry any sense. Consider a liquid contained in a round bottomed flask fitted with a long narrow stem as shown in fig. Let the initial level of the liquid be X. When it is heated the level falls initially to Y.
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However, after sometime, the liquid level eventually rises to Z. The entire phenomenon can be understood as follows: Upon being heated, the container gets heated first and hence expands. As a result, the capacity of the flask increases and hence the liquid level falls. After sometime, the heat gets conducted from the vessel to the liquid and hence liquid also expands thereby rising its level eventually to Z. Since, the volume expansivity of liquids, in general, are far more than that of solids, so the level Z will be above the level X. Effect of temperature on density When a solid or liquid is heated, it expands, with mass remaining constant. Density being the ratio of mass to volume, it decreases. Thus, if V0 and Vt be the respective volumes of a substance at 0°C and t°C and if the corresponding values of densities be 0 and t, then the mass m of the substance is given by m = V00 = Vtt But Vt = V0 (1 + t), so t = 0 (1 + t)–1 Ex.
[Sol.
The volume of a glass vessel filled with mercury is 500 cc, at 25°C. What volume of mercury will overflow at 45°C? the coefficients of volume expansion of mercury and glass are 1.8 × 10–4 / °C and 9.0 × 10–6 / °C respectively. The volume of mercury overflowing will be the expansion of mercury relative to the glass vessel (i.e., apparent expansion). Now, since (V)a = (V)r – (V)c Apparent expansion (V)a will be (V)a = V11T – VCCT = 500 cc (180 – 9) × = 1.71 cc Thus, 1.71 cc of mercury overflows.
10 6 (45 – 25)°C C
]
Dulong and Petit’s Method It is a special method of determining the cubical expansivity of a liquid without the intervention of the container. fig. shows the experimental arrangement consisting of a glass U tube filled with the experimental liquid to the same height in the two limbs. Both the limbs are enclosed in two different jackets, maintained at steam point (100°C) and ice point (0°C) by means of steam and ice cold water circulated in the two jackets respectively.
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The horizontal portion of the U tube is covered with a blotting paper continuously soaked in cold water, to avoid heat flow from the liquid in the limb at a higher temperature of that at a lower temperature. The two jackets contain accurate thermometers to yield steady readings, the levels of the liquid in two limbs are recorded (say h1 and h2). Since, the base of the liquid columns in the two limbs are at the same horizontal level, pressures, at those points with be the same. If 0 and 100 be the densities of the liquid at 0°C and 100°C and P0 the atmospheric pressure, then P0 + h10g = h2100g + 0 or h10 = h2100
0 100 = 1 .100
So,
h2 = h1 (1 + 100 )
or
h 2 h1 = 100 h 1
The experiment can be carried, even with temperatures other than the ice and the steam point. If T1°C and T2°C be the temperatures of the liquid columns in the two limbs, corresponding to heights h1 and h2 respectively, then P0 + h11g = P0 + h22g or h11 = h22
Ex.
or
h1 0 h 2 0 = 1 T1 1 T2
or
h2 1 T2 = h1 1 T1
or
h 2 h1 = (h T h T ) 1 2 2 1
]
An automobile radiator A small plastic container, called the coolant reservoir, catches the radiator fluid that overflows when an automobile engine becomes hot (see figure).
°
°
°
°
Radiator
An automobile radiator and a coolant reservoir for catching the overflow from the radiator The radiator is made of copper, and coolant has a coefficient of volume expansion of = 4.10 × 10–4 (C°)–1. If the radiator is filled to its 15-quart capacity when the engine is “cold” (6.0 °C), how much overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92°C? Reasoning When the temperature increases, both the coolant and radiator expand. If they were to expand by the same amount, there would be no overflow. However, the liquid coolant expands more than the radiator, and the amount of overflow is the amount of coolant expansion minus the expansion of the radiator cavity. Page # [20]
Sol.
When the temperature increases by 86°C, the coolant expands by an amount V = V0T = [4.10 × 10–4 (C°)] (15 quarts) (86 C°) = 0.53 quarts The volume of the radiator cavity expands as if it were filled with copper [ = 51 × 10–6 (C°)–1. The expansion of the radiator cavity is V = V0T = [51 × 10–6 (C°)–1] (15 quarts) (86 C°) = 0.066 quarts The amount of coolant overflow is 0.53 quarts – 0.066 quarts = 0.46 quarts Anomolous expansion of water While most substances expand when heated, a few do not. For instant, if water at 0°C is heated, its volume decreases until the temperature reaches 4°C. Above 4 °C water behaves normally, and its volume increases as the temperature increases. Because a given mass of water has a minimum volume at 4°C, the density (mass per unit volume) of water is greatest at 4 °C, as figure shows.
Density, kg/m3
Maximum density at 4°C
1000.0 999.9 999.8 999.7 999.6 0
2 4 6 8 Temperature, °C
10
The density of water in the temperature range from 0 to 10°C. Water has a maximum density of 999.973kg/m3 at 4°C. (This value for the density is equivalent to the often quoted density of 1.000 grams per milliliter) When the air temperature drops, the surface layer of water is chilled. As the temperature of the surface layer drops toward 4°C, this layer becomes more dense than the warmer water below. The denser water sinks and pushes up the deeper and warmer water, which in turn is chilled at the surface. This process continues until the temperature of the entire lake reaches 4°C. Further cooling of the surface water below 4°C makes it less dense than the deeper layers ; consequently, the surface layer does not sink but stays on top. Continued cooling of the top layer to 0°C leads to the formation of ice that floats on the water, because ice has a smaller density than water at any temperature. Below the ice, however, the water temperature remains above 0°C. The sheet of ice acts as an insulator that reduces the loss of heat from the lake, especially if the ice is covered with a blanket of snow, which is also an insulator. As a result, lakes usually do not freeze solid, even during prolonged cold spells, so fish and other aquatic life can survive.
{Home Work - HCV : Chapter -23
Ex 10 - 34, Question & Short Answer Leave 4, 5 Obj. - I Leave 1 to 4 Obj. - II Leave 4 }
Note : For any suggestion or correction please contact Amit Gupta or give it in computer room.
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