 Welcome to the sixth lecture of cryogenic engineering for NPTEL program. The title of this is continuing from the earlier lecture which is properties of materials at low temperature. Here, we are considering various properties that are mechanical properties, thermal properties, electrical properties and magnetic properties. We have talked about this in the last lecture out of which the mechanical and the thermal properties are basically meant for mechanical engineers and this is what the course is meant for mechanical engineers. And therefore, we will talk about these properties, mechanical properties and thermal properties at low temperature in detail. At the same time, I will talk about electrical and magnetic property. I will touch upon these properties and I will not go into the details of these properties. And as far as electrical and magnetic properties are considered at low temperature, we will talk more about superconductivity. This S c is basically short form of superconductivity. So, we will talk more about mechanical and thermal in detail for mechanical engineers, while I will just give some information about these two property giving more stress on superconductivity as such. My earlier lecture introduced you different material properties. We talked about structures of metals and plastic. We talked about B C C, H C C, X C C C etcetera all those materials and their relevance to the mechanical properties. We also talked about the stress strength relationship which are very very important in order to understand mechanical behavior of a material at any temperature. We talked about in detail about the mechanical properties of metals at low temperature. At the same time, I touched upon the mechanical properties of plastics at low temperature because plastics also play an important role in cryogenics. In today's lecture, I will continue with the material properties of low temperature. I will give a small introduction as to what exactly I am going to talk about in today's lecture. Then the thermal properties. The essence of today's lecture of the main focus of today's lecture will be on thermal properties and thermal property variations at low temperature. Also, I will talk in brief about electrical and magnetic properties. Now, as far as the properties of the materials at low temperature are considered, the properties of the material change when cooled to cryogenic temperature and we have seen a small video. We showed to you how the properties of the material change at low temperature. We know that the electrical resistance of a conductor decreases as the temperature decreases. In fact, the electrical resistance decreases as the temperature decreases. So, as you go on reducing the temperature, the conductivity increases. In other word, electrical resistance decreases. At the same time, we know that if the material's temperature is increased, it expands. Similarly, in cryogenics, if the material's temperature is decreased, the material will shrink and therefore, the shrinkage of material occurs when cooled to low temperature and this is very important consideration and this is what we will see in this particular lecture. At the same time, we understand that the system is cooled on faster at low temperatures due to decrease in the specific heat. If I want to cool down the system at room temperature, the system will take some more time as compared to what it would take at lower and low temperature, essentially because there will be decrease in specific heat at low temperatures and this is an aspect which has to consider while cooling different materials at low temperature. With these small examples, we have very general examples we talked about. We know that the study of properties of material at low temperature is necessary for the proper design because every cryogenic design, every cryostat design, every cryogenic device has to understand how the materials behave at low temperature and their behavior has to taken into consideration while designing that particular device. Now, what are different material properties we are going to talk about under the thermal properties. So, under thermal properties, I will talk about essentially three properties. One is thermal expansion or contraction. In cryogenics, we will come across more contraction if we are reducing temperature while if we are increasing temperature, we will get expansion. Second property is specific heats of solids because specific heat is a very important function of temperature and this we try to understand in this particular lecture. Important thing is thermal conductivity. So, we will study these three properties and the variations of these properties at low temperature in detail. So, let us first come to thermal expansion. What is thermal expansion? Thermal expansion or contraction is nothing but reduction in the dimensions of a material occurring when cooled to low temperature. Simple, we know this that if we heat the material, its dimension increases. Similarly, if the material is subjected to very very low temperature, its shrinks and that has to be considered in the design of a cryostat or a cryogenic device. Now, I just show a small schematic or a small animation here where we have got a material A and material B and these two materials are joined together by a brazing or a welding or whatever process of joining we use over here and this joining naturally is then at 300 Kelvin I mean at room temperature. 300 Kelvin is always treated as room temperature. Now, this joint when subjected to low temperature will look like this. Now, why does it look like this? It shows that material A has shrunk also material B has shrunk and you can also understand that the joint which was perfect at room temperature. Now, that joint is not perfect here and the joint will start leaking and why does it leak? It leaked because both the material shrunk and the joint therefore, gave way to whatever cryogen or whatever gas flowing beneath this. That means at 300 Kelvin the joint was perfect at 80 k the joint has become imperfect the joint has started leaking. Now, this is a very important thing. This is where cryogen is comes into picture. This is where thermal engineer comes into picture. This is the contraction of A and this is the contraction of B. Now, any designer who knows this will have to consider the shrinkage of material A and shrinkage of material B before he starts joining mechanism and this requires some of the calculations to be done. Some properties of material have to be known the shrinkage is a function of temperature. So, one should know how much shrinkage it comes across? How much material A shrinks or how much material B shrinks? Does material shrink more than A or more than B? That you have to understand and those things have to be taken into consideration while designing a cryostat or any machine. So, let us see what is that parameter which defines the shrinkage. The shrinkage is now defined by the linear coefficient of thermal expansion when we are talking about one dimension or let us say length, width or height etcetera. We can define the shrinkage by lambda t and this lambda t is defined as del L by L upon delta t. What does it mean? It means that the fractional change in length that means if L is a length then del L is a contraction or expansion per unit change in temperature while the stress is constant. So, what is del L by L for a given material per unit change in temperature or per degree centigrade or per Kelvin? That is a very inherent property or a characteristic property of any material. So, lambda t is a specific property of a material and it depends on what is lambda t at a particular temperature because lambda t at room temperature could be different that means the fractional change in length per unit change in temperature could be different at 300 Kelvin at room temperature, it could be different at 150 Kelvin, it could be different at 80 Kelvin. So, this behavior, this variation of lambda t or what is lambda t at room temperature, what is lambda t as 200 Kelvin temperature have to be known before we go for further design of any joint. So, this is a very important thing because some additional stresses can get generated because of contraction and expansion of the joints. The unit of lambda t from this definition is per Kelvin which is k inverse or 1 by k whatever way you want to write it could be written sometimes it is written as even per Kelvin per k. Similarly, now this was linear coefficient of thermal expansion. Similarly, we can have volumetric coefficient of thermal expansion beta and is a fractional change in volume per unit change in temperature while the pressure is constant. So, you can have one in length, breadth and height which will give a volumetric change in thermal expansion. Volumetric change occurring in all the three dimensions and for isotropic materials where lambda t in all the three directions is same, we can have beta is equal to 3 lambda t that is understandable. Now, this variation, this particular graph gives the variation of lambda t with respect to temperature. So, we can understand now that lambda t is basically decreasing trend for every curve. This curves are meant for different materials. This is for aluminum, stainless steel, nickel, carbon steel and titanium as shown over here. And what you can see from here that lambda t is basically decreasing with the decrease of temperature. So, one thing is sure that the variation of lambda t for the few commonly used material in all the materials, we see that the coefficient of thermal expansion is decreasing with the decrease in temperature. So, this has to be understood. Also, what you can see is at 300 Kelvin, lambda t's values are much higher as compared to let us say 80 Kelvin temperature. That means, at room temperature the shrinkages are much higher, delta L by L per degrees centigrade are much higher as compared to very low temperature. This aspect also has to be considered while designing the machine. So, you can see below 80 Kelvin the lambda t value is just 10, 8 or 4 while at 300 it is of the order of 24, 25 etcetera. So, most of the contraction we can see that it happens from 300 to 80 Kelvin while below 80 Kelvin the contractions are that way negligible. Now, why does this contraction happen? In order to understand that we can understand that we can see from this figure that the molecular internal energy variations versus the intermolecular distance which is given over here. This figure gives the variation of molecular internal energy versus intermolecular distance which is given by R. What you can see here is at 0 point that is at temperature equal to 0, R 0 has got particular value. What you can see from here this particular curve shows that as the temperature increases this is the oscillation of the atoms in a particular lattice and this oscillation amplitude has increased. As the temperature increases this oscillation amplitude increases and you can see here as the temperature increases further this oscillation amplitude has further increased. Also what you can see from here that the equilibrium spacing curve which basically denotes the intermolecular distance the curve is tilting towards right which shows that this is not a very symmetric curve. This what we see from here the equilibrium spacing basically depicts the mean position of the atoms about which it oscillates. With the rise in temperature the increased thermal agitation leads to increased intermolecular distance which is what we just saw and the energy curve is asymmetric about the point R 0 stating that the atomic vibration is asymmetric. So, we can understand that as the temperature increased the curve is leading towards right. It means that they are not symmetric while when the temperature increases from T 1 to T 2 the oscillation increased while if it goes from T 2 to some additional T 3 value the oscillation will further increase and that increase will be much more than the ratio of this temperature T 3 to T 2 or T 2 to T 1. As soon as you understand this we can understand that the lambda value are changing and the lambda values are more at higher temperature than at low temperature. This is what the conclusion could be drawn from this the rate of increase of intermolecular distance increases with the increase in temperature. So, as you increase from T 1 to T 2 or from T 2 to T 3 smallest increase at T 2 to T 3 can lead to much more increase in the intermolecular distance which basically means that the lambda value is much more here as compared to what it is over here and this basically offers the explanation for the curve which we earlier saw that lambda values at 300 Kelvin are much higher as compared to what they are at let us say 80 Kelvin or 50 Kelvin. Hence, the coefficient of thermal expansion lambda t increases with the increasing temperature and this is basically the reason why material expands when heated and why do they contract when they are cooled. There is one more parameter which has to be understood called mean linear thermal expansion which is delta L by L 0. Now, this parameter also gives you understanding as to what exactly happens when the temperature changes happen from let us say from 0 Kelvin to a particular temperature T. So, if I define mean linear thermal expansion as del L by L 0 which is L t minus L 0 divided by L 0 that means length at any temperature T minus L 0 that is the length at 0 Kelvin divided by L 0. So, this curve basically derived from the values of lambda t for different materials. So, you can see from here that aluminum has got more delta L by L as compared to copper as compared to stainless steel as compared to nickel. So, here we can understand that let us say for temperature equal to 280 Kelvin delta L by L at 280 Kelvin will be around 375 that means if we change the temperature from 0 Kelvin to 280 Kelvin the length increases around 375 or delta L by L into L 0 into 10 to the power 5 is equal to 375 which means around 3.75 millimeter if we have got a 1 meter length at 0 Kelvin. So, this curve gives you the change in length for a given change of temperature from 0 Kelvin alright. Again from this curve you can see that below 80 Kelvin the delta L by L value are much smaller as compared to what they are at higher and higher temperatures. The slope of the curve is very high up to 80 Kelvin and thereafter it flattens. The mean linear thermal expansion can also be evaluated between two different temperature. Earlier we saw that mean linear thermal expansion was with respect to 0 Kelvin, but now if I have got two temperature into consideration like L T 1 and L T 2 are the lengths of the temperature at T 1 and T 2 respectively then the change in length if we change the temperature from 280 to 80 Kelvin or 80 to 280 Kelvin whatever then we can compute those two temperatures by this formula. So, in this case now delta L by L 0 will be L T 1 minus L T 2 upon L 0. So, if I get the value of L T 1 by L 0 which is let us say this and L T 2 by L 0 which is value this and if I get the difference between these two values then what I get is the delta L value if I change the temperature from T 1 to T 2. Now, this is a very important calculation which one has to do if one uses liquid nitrogen at 77 Kelvin from room temperature how much that particular part will shrink can be computed over here and this is very important to understand what will be the stresses generated how the joint will behave at lower and lower temperature this is very important and with this formula one can calculate the delta L value between two temperature changes. So, here you can see that for different materials we have calculated the temperature of the stainless steel rod decrease from 280 Kelvin to 80 Kelvin we find from here this is stainless steel over here that d L equal to 2.5 non millimeter if L equal to 1 meter at 0 Kelvin and from 80 Kelvin to 20 Kelvin d L is only 0.13 millimeter telling you that this indicates to you that maximum shrinkage again has occurred up to 80 Kelvin while below 80 Kelvin it is only 0.13. One can compare this with the value of lambda T which we earlier saw from here you can understand that for aluminum the shrinkage is much higher which is clear from this particular portion that values for aluminum are much higher then comes stainless steel then comes nickel and then comes carbon steel. So, basically the shrinkages go in order of the lambda T values and if one wants to compute the exact shrinkage one can compute by this formulae and these are very very important in order to understand how much shrinkage would occur and therefore what kind of jointing mechanism you should device or how much stresses will get generated if this particular rod is subjected to low temperature. With this now I will switch to the next property which is specific heat of solids what is specific heat this is also a very important property it is the energy required to change the temperature of a unit mass of substance by 1 degree centigrade holding the volume or pressure as constant. So, basically how much amount of energy is required to change let us say 1 kg substance by 1 degree centigrade which talks about its C p value or C v value by thermodynamics you know that C v is nothing but d u by d t that means the change of internal energy per degree centigrade temperature. Similarly, the C p value or the specific heat at constant pressure is nothing but variation of enthalpy T h by d t change of enthalpy per degree temperature change. Now, in order to understand how to calculate C v and C p or how the C v and C p change at low temperature I am going to consider only solids over here because that is the material normally we come across in calculations. Now, lot of things evolved from right earlier from 1911 and in 1911 Dolong and Pettit observed that the heat capacity of the solid are independent of temperature. See that time the low temperature availability was not so much and Dolong and Pettit came to the conclusion that the C p values or the C v values for solids are independent of temperature why because each lattice point absorbs same energy as every other lattice point. So, that means all the lattice point behave in the same way they absorb the same amount of energy and therefore, there is no dependence of amount of energy absorbed over temperature and by equipartition of energy they calculated that u is equal to 3 r t and therefore, d u by d t is equal to 3 r which is C v is equal to 3 r and therefore, they found that d u by d t is nothing but a constant figure. So, whatever happens even if the temperature decreases or increases C v is equal to 3 r value this is what we observed by Dolong Pettit's rule. The next observation which went to Einstein where the quantization came into picture and Einstein treated the solid as a system of simple harmonic oscillators. Now, you know that all the atoms in the molecule oscillate what Einstein observed was that all these oscillations are of the same frequency. So, all the lattice movement that is happening in the lattice all the atoms there which are vibrating they are all vibrating at the same frequency which is not a very realistic observation then, but again it was one of the evaluation that happened over Dolong Pettit's observation. However, the next scientist who came into picture was Debye. Now, Debye treated the solid as an infinite elastic continuum and considered all the possible standing in the material because there are lots of different vibrations that are coming into picture. What are different frequencies coming into picture different oscillating frequencies all the molecules are not oscillating really at the same frequency but, they have got different frequencies. And what Debye concluded is a parabolic frequency distribution exists he derived a parabolic frequency distribution was derived for the atoms vibrating in lattice and he considered all this frequency distribution in order to calculate C p and C v. So, he presented a model to compute lattice hit capacity per mole which accounts for all the vibration frequency for all the lattice points and this is very important thing because this comes very close to the realistic picture of a lattice. So, what he did was the Debye model gives the following expression with lot of calculation he came out with the expression which gives lattice hit capacity per mole and expression is like this 9 R T upon theta d and theta d is nothing but, characteristic Debye temperature and then this expression takes care of the all the lattice vibration frequencies all the oscillating frequencies in the lattice. And he again simplified this by having this expression as d of T by theta d which is called as Debye function. So, this is the most important thing that whatever complicated part was there in order to consider all the parabolic frequency distribution he indicated or he calculated that as a Debye function and one can get the value of Debye function. And therefore, if we know the Debye function at a particular temperature we can compute the C v at those temperatures right. In this particular expression x is a dimensionless variable and this x basically nothing but, h view upon k T which is a Planck's constant into frequency Boltzmann's constant k and temperature. So, from here you can understand that x considered all the frequencies of vibration. In this equation in the equation which is over here the only value of theta d is going to change and what is this theta d is basically nothing but, Debye characteristic temperature. So, for different materials now it will have a characteristic Debye temperature and if I know this Debye temperature I can compute the value of C v provided I know the Debye function values all right. So, the whole thing become very simple one can compute the effect of temperature variation on the values of C v. From this expression we can understand that at very high temperature that means, T greater than 2 times theta d the specific it obtained from the above equation approaches to 3 r values and this is nothing but, what we got earlier by Dolong Pettit's value. So, at higher temperatures at very high temperatures at room temperatures and above what we will have is the C v is equal to constant values which is equal to 3 r and at low temperatures where T is less than theta d by 12 this will come into a very low temperature category. The Debye function here in this case approaches a constant value and at d 0 what we have is a 4 pi to the power 4 by 5 is constant value. If I put the Debye value over here particular C v value will be function of only temperature now in this case while all other things are constant. You can see here now at lower temperature the C v variation is T cube directly perpendicular to T cube all right and this is a very important conclusion Debye came up with that the variation of a cubic equation in absolute temperature at very low temperature C v is directly proportional to absolute temperature to the power 3. So, at lower and lower temperature you can imagine this behavior of C v will be and this is the curve which basically gives the value of C v by r with T by theta d. So, if I know the T and if I know theta d theta d is a characteristic temperature for any material. So, what you can see is for higher temperatures where T is more than 2 times theta d or 3 times theta d the C v by r is almost constant which is actually equal to 3 times r while at lower and lower temperatures what you see is C v by r is coming down with the power of 3 as a temperature coming down and therefore, you can see the steep decrease at a lower temperature side and this is a very important curve in order to calculate the value of C v. So, in general what you can see from here is that the specific heat decreases with the decrease in temperature all right which is obvious from this particular figure. Now, this table gives you different values of the theta d for different materials. So, you got a aluminum 390, you got a beryllium, you got a calcium copper etcetera. If I know the theta d values for different temperature for different materials then I can calculate T by theta d and then I can go back to this curve I can compute the T by theta d and calculate C v of that particular material it is as simple as that. So, if I know theta d I can compute T by theta d and then I can correspondingly compute the value of C v in that case. So, if I want to calculate the value of C v what I am saying I am repeating the same thing the calculation of C v for a particular material at a particular temperature T inverse following procedure refer to the table and find out the value of theta d calculate T by theta d and interpolate the value on the graph to obtain C v by R. C v by R is known just multiply C v by R by R value where R is the specific gas constant and then what you get is the value of C v. If the value of T by theta d is less than 1 by 12 then we know that we have got one more expression which has been given earlier we can use that correlation we use the value of T by theta d and compute the value of C v directly in that case all right. So, this is how we can calculate the value of specific heat at lower and lower temperature if one knows the value of theta d for that particular material. The next property which I am going to talk about now is the very important property again thermal conductivity in solids. Now, why do we require the knowledge of thermal conductivity in solids at low temperature is because of some examples I am trying to give you. So, in a cryostat the solid members could be metal or non metal could be stainless steel could be copper could be plastic nylon etcetera and this material conduct heat from high temperature to low temperature. Sometimes we have got some suspension member in order to support a cryostat which will definitely take heat from room temperature and take it to the low temperature. So, for these members the thermal conductivity K T should be as low as possible otherwise they carry lot of heat from room temperature to the cold temperature and therefore, the cryogen there will be more loss of cryogen or some additional cooling effect has to be used for this and therefore, we have to see that the conductivity of this particular material is very very low we have to see that thing. On the other hand I will give the other extreme example wherein you want to now transferred cold because you want to cool something because lot of cold has been generated in a system let us say in a cryocooler and we want to cool a sample or we want to cool a gas and therefore, we want to have a material which has got very high conductivity at low temperature and in this case I would like to achieve base heat transfer of the cold generated and therefore, copper can be used as a medium due to its high thermal conductivity. So, I have to select a material in such a way that where I want minimum losses to be coming on a system then I should use low thermal conductivity material like nylon like stainless steel, but wherever I want that the heat transfer should be as high as possible therefore, I should use the material which has got high thermal conductivity like copper like aluminum etc. The thermal conductivity in solid let us say KT is a property of a material which indicates its ability to conduct heat. Now, this particular curve this particular graph shows the thermal conductivity variations at lower and lower temperature. You can see two types of curve one is a curve which is going up and then coming down which is for pure material or pure copper in this case. Well you can see other materials like aluminum, carbon steel, stainless steel we show that as the temperature is coming down the temperature below let us say 200 Kelvin is coming down continuously like this. So, these are basically alloys or impure material we can call them well this is the pure material, but in general you can understand for these materials KT decreases with the decrease in temperature. However, as I just indicated to you earlier for pure materials the variation is slightly different or one can say even quite different from as compared to the impure materials or alloys. So, this is very important to understand what exactly happens over here, what exactly happens over here and I will just try to give you some explanation for these two different curves. For pure materials the for any material for that matter the electron and phonon motion cause heat conduction right. Basically thermal conductivity is because of the motion of electrons and phonon. Phonon is basically the lattice vibration and these two contribute to the heat conduction. Now, the contribution of the electron motion to heat conduction is predominant above liquid nitrogen temperature. As you know that liquid nitrogen temperature is a dividing line in cryogenics and above liquid nitrogen temperature or let us say 77 to 80 Kelvin above this the electron motion of heat conduction is predominant as compared to phonon motion alright. So, at temperatures below l n 2 below liquid nitrogen temperature phonon motion is dominant. So, you got a two variations above liquid nitrogen temperature what you have is a electron motion contribution is more while below it is not electron motion but phonon motion is predominant at lower temperature and this is basically what makes the difference in the behavior as you see in pure material. Now, the thermal conduction depends on the product of in electronic specific heat and mean free path. So, if I say q which is a conduction is basically directly proportional to two parameters that is electronic specific heat and mean free path. Now, this product is constant above l n 2 alright. So, what is happening above l n 2 is as you come down the temperature up to let us say 80 Kelvin from room temperature the electron specific heat is decreasing while the mean free path is increasing. The electron mean free path is increasing and therefore, somehow the product of this specific heat electronic specific heat and mean free path is remaining constant above liquid nitrogen temperature which makes that the thermal conductivity k t is remaining constant above liquid nitrogen temperature right. What is happening below as I said below the liquid nitrogen temperature it is not the electronic specific heat which is predominant but the phonon specific heat is predominant. So, as the temperature is lowered the phonon contribution increases and the k t is varying as 1 by t square. So, you can see that if the temperature is decreasing the phonon contribution is increasing in this case which was not. So, as far as the electronic contribution was considered also as we know earlier that as the temperature is coming down the mean free path is increasing alright. So, here what you see now at below l n 2 temperature the phonon contribution mean free path also increasing the value of k t. So, at lower temperature from let us say below l n 2 the phonon contribution increase plus the mean free path also has increased which leads to increase in thermal conductivity for pure material. What this will happen only up to a particular temperature what happens at this temperature? At this temperature the thermal conductivity reaches a very high value until the mean free path of the electrons equal to the dimensions of the test specimen. So, naturally the mean free path has got some limitation and it increases to a maximum to a value which is equal to the maximum dimension of the test specimen alright. So, beyond which now it cannot increase that means it has reached a constant value and therefore, below this temperature now below this temperature the conductivity starts coming down at this point or it reaches a maximum over here and because it has reached a maximum value when this condition is reach that it has reached the maximum value the surface of the material exhibits a resistance to the motion of electrons alright. So, basically the resistance is offered by this material to the flow of electrons and flow of phonons and therefore, it causes the value of k t to come down below low temperature below 10 Kelvin region etcetera. So, this explains why you got a constant kind of a curve up to let us say 80 Kelvin and below 80 Kelvin it goes very high up to this point because of the phonon contribution and because of the electronic mean free path increase and then it starts coming down because now electron mean path has reached to a maximum value and the surface has started offering resistance to the flow of electrons. This is a very funny behavior this is a very typical behavior as far as the pure materials are considered. As far as the impure and alloy metals are considered we can say that the electron and phonon motion in this case are of same magnitude. So, the contribution which is coming from electron and phonon both is almost of the same magnitude in impure and alloy metals alright. So, below a particular temperature phonon is predominant above a particular temperature electron is from predominant that aspect is not existing in this case both of them contribute equally as far as the impure materials or the alloy materials are considered. What happens for the decrease in the thermal conductivity in these materials then the impure metals have got imperfection and that is why they are called as impure materials and they have got imperfections like grain boundaries and dislocation you know about this we have talked about this in the earlier lecture. So, you have got imperfections like grain boundaries and dislocations and this basically are kind of hindrances to the motion of electrons right. What happens the electrons strike these imperfection and they get scattered alright. So, an additional scattering of electrons occur due to grain boundaries and dislocations. Now, this scattering is proportional to T cube for grain boundaries and T square for dislocation alright at very low temperature what it means is as the temperature comes down the scattering become less and less and the less is T cube proportional and T square proportional if we talk about a particular dislocation of grain boundary or dislocation respectively which means that we are getting an additional advantage of scattering because of which electrons were conducting heat from one point to another, but this is not happening at low temperature because the scattering has reduced drastically at low temperature. So, at low temperatures scattering decreases as a result K T decreases with the decreasing temperature in impure metals and alloys. I am sure it is clear to you that in this particular cases the scattering phenomena decreases at low temperature and therefore, the thermal conductivity in this case decreases at low temperature. These materials do not exhibit any high maxima like that of pure metal. So, what we saw in a pure material we do not see in this particular case thermal conductivity integrals this important parameter which has to be understood while calculating the thermal conductivity loss at low temperature. As we found that the thermal conductivity K T is a strong function of temperature the term K T dT is very important. So, we have just seen that the K thermal conductivity is a strong function of temperature. Due to this variation in K T the calculation of heat transfer Q can be very complicated. If I want to calculate the conduction loss between 300 Kelvin to 80 Kelvin I have to consider the K variation with temperature and then calculate for each temperature in order to know how much what is the loss of conductivity between these two temperatures. And therefore, a simple method is proposed in order to simplify the calculation of Q. This method uses integral K dT as a parameter is called as thermal conductivity integral which basically sums up the effect of variation of K T with respect to temperature change alright. So, K T dT is a parameter which is directly offered for a particular material and therefore, one can get the what is the K dT parameter in a particular given temperature interval and this benefits a lot by calculating the loss due to thermal conduction. How we can see now? The Fourier's law of heat conduction is given by the following mathematical expression which you know Q is equal to minus K into A into dT by dx alright. Now, this we can club as K T and dT can be clubbed together while A x upon dx can be clubbed together and if we do that thing what we get is this. So, Q is equal to now here minus A x upon dx into K T by dT. In this method Q is expressed as below now I am just representing this Q in this simple format minus G into theta 2 by theta 1. Now, this G is nothing but this parameter A x upon dx which almost remains constant for a regular physical material and K T dT is nothing but theta 2 minus theta 1 that is temperature let us say T 2 and T 1 it talks about these two parameters alright. So, A x and d x is basically represented by G while K T dT is given by theta 2 minus theta 1 where theta 1 and theta 2 are expressed as thermal conductivity integrals. So, theta 2 is a thermal conductivity integral at temperature T 2 while theta 1 is a thermal conductivity integral at T 1. So, K dT now is taken as integrals called thermal conductivity integrals evaluated with respect to datum temperature. So, we found that theta 1 is nothing but what is the K T dT from T d or any datum temperature to temperature of my interest let us say T 1 and similarly theta 2 will be from T d to T 2. The datum line should be same in both the cases and this datum line various books gives various datum lines could be 0 Kelvin or 4.2 Kelvin alright. So, if A c s that is area of cross and is constant G is defined as A c s by L which is what we saw and from this now what we see is a curve which gives variation of K dT versus temperature. So, if I talk about 100 Kelvin I have got K dT value for a particular material as this which means that from 0 Kelvin to 100 Kelvin I have got K dT value up to this point and let us say at 10 Kelvin I have got K dT value as this and if I take a difference of this two I can calculate the loss of conduction how much loss we have because of the thermal conductivity from 100 Kelvin to 10 Kelvin. In the calculations the actual temperature distribution is not required but what you require is only the end point temperature I do not have to worry what happens from 100 to 10 Kelvin I do not have to worry about 90, 95, 85 exactly like I just know the value of K dT at 100 Kelvin I need to know the value of 10 Kelvin if I know these two values I can straight away calculate the amount of Q that will be loss because of the thermal conduction. So, here this technique is widely used in the analysis of heat leaks if the day term temperature is taken at 0 Kelvin and the two ends of a specimen are maintained at 100 Kelvin and 10 Kelvin respectively then the K dT integrals are given by if I want to calculate now amount of loss is occurring because of temperature change from 100 Kelvin to 2 Kelvin for a particular material if I am talking about aluminum, phosphodon or stainless steel whatever then what I am going to do is find a value of K dT for a particular material it is a phosphor bronze at 100 Kelvin which could be around this minus at value of K dT at 10 Kelvin and the difference between the two is nothing but the heat loss due to conduction in a temperature region of 100 to 10 Kelvin temperature what I need to know only one value here one value here and the difference between the two will give me K dT multiplied by G is basically going to give me the amount of heat lost because of thermal conduction with these three properties what we just talked about is thermal expansion we talk about the specific heat capacity we talked about the thermal conductivity with these three thermal properties I will come to now electric and magnetic property I will touch in very brief in this case the electrical conductivity as you know it is defined as the electric current per unit cross section area divided by voltage gradient in the direction of the current flow. So, we know that V upon I is nothing but resistance I am talking about the reciprocal of that that is electric current per unit cross area divided by voltage gradient in the direction of the current flow this is what electrical conductivity is given as the electrical resistivity is opposed to that alright. So, electrical resistivity is it is reciprocal of electrical conductivity as you know that decreasing the temperature decreases the vibration energy of the ions as you know that as the temperature decreases the energy decreases and therefore the vibration energy of the ions also decreases this result in smaller interference electron motion these ions these are positive charge actually ions which are basically are the resistance that is offered to the motion of electron motion electron motion creates a current. So, at lower and lower temperature the resistance offered by these ions is less and less and therefore the electrical resistivity in this case decreases. Therefore, electrical conductive the metallical conductor increases at low temperature and it is a very simple physical explanation to understand why at low temperature electrical conductivity of the metallic conductor increases at low temperature. Now, this particular curve gives the electrical resistivity ratio with temperature for different materials like aluminum, iron and copper and what you define electrical resistivity ratio as is defined over here the E r r is defined as r T upon r 273 that means r T is at any temperature T and r 273 is nothing but 273 Kelvin at that is 0 degree centigrade. So, you can understand that r T by r 273 is equal to 1 for all the materials and it is this point the variation of electrical resistivity ratio for some commonly used material is as shown over here and what you can see now r T upon r 273 decreases for all the materials over here. However, the decreases are different with different temperatures with different materials. So, what you can see for copper r T at any temperature let us say at 100 Kelvin r T upon r 273 for copper is much smaller as compared to what it is for aluminum. What does it mean that r T at 100 Kelvin r 100 is much smaller value as compared to r 100 for aluminum which means conductivity of copper at 100 Kelvin is much higher as compared to what it is for iron or what it is for aluminum which is what we know alright. So, this is very important to understand that how this resistivity varies at low temperature or how this ratio varies at low temperature. Also important to understand that the electrical and thermal conductivity are related by Wiedemann Frehn's expression. What does it do? Wiedemann and Frehn's expression basically finds a relationship between thermal conductivity and electrical conductivity alright. So, K T and K E are nothing but thermal conductivity and electrical conductivity. So, K T upon K E into T is equal to 1 by 3 pi K upon e to the power 2. You can see that this particular parameter is almost constant because K is a Boltzmann constant is the charge per electron and you can see that this is a completely constant value which will not change with temperature. If suppose the T comes on this side then you can understand that K T upon K E is a function of temperature only. It means that the ratio of K T upon K E is a product of a constant which is this and absolute temperature and is represented like this. So, K T upon K E is equal to A T. It means that as the temperature gets lowered if we are reducing the value of T the K T upon K E also gets reduced. It means that K T gets reduced or K E increases. The two possibilities we know that the thermal conductivity decreases at lower temperature. We also know that electrical conductivity increases at low temperature because of decrease in thermal conductivity and because of increase in K E at lower temperature K T by K E they move in such a way that at a particular temperature K T by K E remain constant or if the variation of temperature is given K T by K E will move according to this particular expression. This finishes whatever I want to talk about with regard to the electrical conductivity. Just to sum up what we have done in this particular lecture is the coefficient of thermal expansion decreases with decreasing temperature. This is what we talked about lambda T. Lambda T was nothing but equal to del L by L divided by delta T. So, what we understand from here is for different materials you have got specific lambda T value and in all those materials as the temperature decreases the lambda T value decreases which means that the material whatever it is it shrinks when temperature gets lower or it expands when the temperature increases. For pure material like copper what we saw K T remains constant above liquid nitrogen temperature. Below liquid nitrogen temperature it reaches a maxima let us say around up to 10 Kelvin and after that below this particular temperature it decreases steadily and we have understood the reasons as to why does it remain constant above liquid nitrogen temperature, why does it go from l n to temperature below up to 10 Kelvin and why does it basically later on decreases below 10 Kelvin to the value 0. Impure materials K T decreases with decrease in temperature and also we understood that the integral K D T is used to calculate q and we also saw how this K D T value is calculated from the curve. The electrical conductivity of the metallic conductor increases at low temperature this is what we understood at the same time the electrical conductivity and thermal conductivity are correlated by Wiedemann Franz law. So, here we summarize whatever we have discussed in today's lecture. Now, you must have realized that there are lot of equations, lot of curves, lot of tables wherein you should be able to calculate the specific heat at particular temperature for a particular material or the shrinkage of a particular material at particular temperature or the loss due to thermal conduction from let us say room temperature to 77 Kelvin or room temperature to 4.2 Kelvin or if the particular rod is attached between 100 Kelvin to 10 Kelvin then how much load it gives on a 10 Kelvin cold that has to be calculated that has to be understood and all these problems in the form of tutorial we will discuss in the next lecture which will be a very important lecture to understand how to calculate these losses where the mathematical relationship between these particular parameters and the tables will come into picture. At the end I have got some self assessment exercises given kindly assess yourself for this lecture very honestly they are all self assessment please go through them try to answer and at the end we have given the answers for those questions. So, that you know whether you have answered them correctly or not thank you very much.