 Hi, I'm Zor. Welcome to Unizor education. I'd like to solve a couple of problems related to heat transfer Heat transfer problems. Now this lecture is part of the course Physics 14's presented on Unizor.com. I do suggest you to watch this lecture from the website because together with the video presentation you will have the text basically like a textbook in front of you with all the formulas, calculations, etc. So it's very beneficial and The site is completely free. There are no advertisements. So you don't even have to Sign in if you don't want to take exams, but if you want to challenge yourself, please do Okay, so let's get to the problems. So I would like to solve the problems in in general with some unknown variables basically describing the physical Objects we are dealing with and then maybe if I have time I will use some concrete numbers to basically do whatever it is If not, the numbers will be in the text part of this lecture just for a calculation purpose So that's not very important thing. What's important thing is to approach the problems in their general sense and understanding. So my first problem is Let's consider You have a room Which is supposed to be of certain temperature so we have a certain temperature of the room and There is an outside world Which has its own temperature? My purpose is to maintain a Concrete difference between outside temperature and inside temperature So for instance, I would like my room to be Let's say 15 degrees of Celsius of Celsius Higher than than the outer temperature. Okay, so this delta is given All right, so based on Temperature outside Knowing the delta we can always Calculate what will be the room temperature? But in any case I would like to be dependent only on this delta So if you have a room at certain temperature, you have outside at certain temperature my purpose is To find out how much heat we are losing because of the difference in temperature Which necessitates if I would like to maintain the constant room temperature to have a source of heat Inside the room of the same amount of heat Which we are losing through Conductivity of the of the wall which separates So I assume that the room is perfectly insulated from all the sides except one which is facing the outside world and To make my life easier. I assume that this wall is just solid Wall of some substance of the of the same thickness and in certain area So all I need this information for so I need the area of the room, which is a I need thickness of the room which is L and I need the thermal conductivity Conductivity Which is K So this is a relatively simple problem Related to heat transfer and it has practical Applications as well. So I know what kind of a Difference in temperature I would like to maintain. I know Where actually the heat is going it's going through this wall which has the area of A and the thickness of the wall is L and it's made of some solid material which has certain Coefficient of conductivity thermal conductivity Okay so to maintain this type of a Condition Temperature here temperatures there, etc. I will introduce a function. So let's consider this is my wall This is a room this is outside and I will introduce The point X which is a difference From let's say from the room to any point inside the wall because the heat actually is Sipping through this wall from this surface to that surface, right? This is Assume let's assume that the room temperature is higher than the outer temperature. So the heat goes this way now I Would like to determine the function What is the temperature at each point? From the inside surface to the outside surface now first of all I I think it's reasonable to assume that considering this is a solid wall Which means the penetration of the heat is exactly the same on each layer of this wall I have to assume that this function is linear Depending on X The thicker the wall Obviously the temperature would be closer and closer to T out and on this surface T of zero is equal to T room and T of L Because the thickness of the wall is L is T out and In between from zero to L my function T of X should be linear and I can determine the coefficients Alpha and beta very easily easily from these two. So if I substitute Zero for X, I will have basically alpha, which is equal to T room And if I substitute L, I will have T out equals alpha, which is T room Plus beta times L, right? from which L is equal to T out minus T room divided by L I mean beta is equal to Well quite frankly I would like everything to be Well in this case beta is negative as you see, but so it's minus delta But if I will define delta the other way around Which probably is better because I'm starting from this point from the room So this is the starting point. This is the ending point. So usually Probably would be better if I will consider delta to be negative in which case I don't have to I Don't have to carry this sign. So the sign is inside Delta is negative in this case All right, so I know basically this so T of X is equal to T room plus T out minus T room divided by L X correct So if X is equal to zero, this is zero I will have T room and if X is equal to L I have this is one L divided by L T room and T room will cancel and I will have T out. So these conditions are Actually satisfied. I don't need them anymore. Now I have my function Temperature as a function Of the distance from the surface facing the room Well, if this is the beginning Then that's how it goes Okay, so I know the temperature. That's good Now if I have Two layers, let's say With different temperatures obviously the heat goes through this and we have the law of Fourier, which basically states that the rate of Heat transfer From one layer to another The rate in terms of per unit of time and per unit of area depends basically only on two things on the temperature and the Conductivity So it's basically equals to minus K times My Derivative of of the of the temperature basically if you start if you start from Instead of differential if you will use Just an increment delta you will increment from X to X plus delta X Which will basically give you the similar formula with delta X and then you divide it both both by by delta X and go to it Limit when delta X is going to zero and you will have this for yellow for yeah Which was discussed in one of the previous lesson lessons Now let's recall. This is the function of X which means this is also a function of X So again, this is how much heat goes Through the unit of surface per unit of time Okay Now why is it minus here? Let's just think about it the heat goes from left to right, which means we are basically Transferring the heat towards this now this Temperature temperature is we are assuming this is cooler. This is warm and this is cooler. So the temperature of X is Going down. So the The derivative is negative. So negative and negative will be positive. So we are losing that much of heat That's what's very important. We are losing this much of heat okay now it's Amount of heat which we are losing per unit of time per unit of area now. What's our area? Well, area is a which means that I Can actually multiply it by a to get Total amount of heat which we are losing Through this area right Now what is the derivative? Well derivative of this function. Well, this is the constant. So it goes out derivative is this Right, it's linear function. So derivative is basically this coefficient, which is Delta by the way So I can say that this is equal to minus K Delta and Now I have to multiply by a Because again, this is per unit of area and I have the whole wall Separating room from the outside world. So I have to multiply by the area of the wall So this is Oh, I'm sorry, it's not Delta. It's Delta divided by L. So it's like this so what Basically, this states is that the loss of the heat per unit of time is Constant you see it doesn't depend on X because this is a linear function and obviously it's a linear function because our wall is Solid and consists of exactly the same material. That's why it's linear if it's different Something something like the layer of insulation then maybe the brick maybe another labor installation or something like this Then obviously this will not be a linear function. So I have kind of Simplified my job assuming this is just a solid material solid wall, but in this case, this is the formula and by the way, I have calculated in one particular case when the wall is made of concrete and we know the the conduction of the Of the cement wall or concrete wall, whatever So we know the Delta. Let's say Delta is inside 20 degrees Celsius. I'm outside is let's say 5 degrees Celsius It's a winter time in in north and semi hemisphere So we have differences with 15 degrees the area of the wall Well, again, let's say 4 by 3 meters like 12 meters square Widths of the wall is let's say 20 centimeters, which is 0.2 in my example and I calculated what's the basically Amount of heat which we are losing per unit of time and my calculations show it's about 540 watts So this means we need a source of heat the heater of this type of a power for the Amount of heat where we are losing through this one wall now, obviously in real practical situations We are not really losing through one particular wall. We might actually have some conductivity to other places Well, maybe not if they are on the same temperature level Usually we have windows in the room which definitely complicated the picture So the transfer through the window should be calculated separately from the transfer from the other parts of the wall Plus the ball is not really a solid uniform material So we probably have non-linear function here. It all depends and it all can be calculated slowly Boringly Unfortunately, but it can be done using exactly the same Strategy all you have to do is basically separate your bigger problem into a set of smaller problem Windows one thing Wall is another thing layer of insulation is the third thing etc So you can calculate very slowly separately each of these guys and then you will still have the final result Which is how much heat you need to warm up this room? Okay, that's my first problem Now my problem number two was actually the same problem with calculation, which I have just Skipped and gave you the result 140 and Now I will go to my third problem Also, I will do it in general case first and I'm not sure I will go into the calculations for specific things So my next problem is related to again very practical situation of let's say you have Hot tea kettle just boil the water You have your drink and then it's basically stinging on the stove and cooling down Slowly cooling down. I would like to know What is exactly the temperature of this tea kettle? As a function of a time as it goes now, obviously you understand from just general intuitive Logic that in the beginning when the kettle is very hot the amount of heat which it emits into the room using the Convection of the air is Probably higher than if it's already cooled down and it's almost the same temperature as the air So I was also kind of thinking about what is the graph of the function Temperature as a function of time well, I was thinking that it should be something like this. So this is basically the level of room temperature and This is the level of temperature of the of the tea kettle So let's say in the beginning it's 100 degrees Celsius Which means we just boiled it and then it's basically slowly Slow go goes down first a little faster and then slower and slower So that's what I was thinking which reminds me obviously the exponential function and I basically want to find out if it's really kind of exponential Well Somebody told me a joke that in the real world everything is exponential and if not exponential is logarithmic Which is inverse function. Okay jokes aside. Let's just try to apply the theory of convection To this particular situation Now I would like to simplify my job again. It's very typical for physicists First they model the real world And then they realize that the model is not actually the same as real world They're trying to complicate it more and more until it may be closer and closer to the real world But right now I would like to really simplify my my work my my my problem down to a very kind of a palatable case Now let's just think about if this is a real tea kettle Well, it has a metal outside. It has some remnants of the water whatever we have not used for tea Which means obviously it's complicated. I mean first you have the convection Towards The air then it's not immediately whatever you lost your your heat from the surface of this tea kettle doesn't immediately actually Go doesn't immediately cool down the inside the water So maybe there is a convection inside of the tea kettle as well. So complicated too complicated We are not considering this my consideration is much much much simpler instead of a tea kettle in the air. I will have a thin square Object of Certain temperature, let's say Tea of tea that would be the temperature of this as the time goes obviously Tea of zero should be given In case of a tea kettle right after it it boiled the water it would be 100 degrees Celcius, but in any case there is some kind of an initial temperature Then around this there is some kind of a substance. I don't care what whatever it is liquid or or gas which basically Is taking the heat out from this Thin square and the temperature of outside is constant. I assume this substance has TS temperature Because the outside is very large infinitely large in my model the outside world is infinitely large Which means whatever the temp whatever the heat this thing emits it It's completely dissipates and doesn't really change this temperature Because if it does it's another complication and I have to take it into account, which is Again, not not very difficult, but I would like to simplify this job my job So we have this infinitely large outside substance which you it which is used like a sink for the heat the heat goes from here and It dissipates immediately using the convection now for instance This is water like ocean right and this is just a hot square of L by L and It has certain mass and now why did I? Specify that it's very thin well because if it's very thin then the Cooling of the surface here or here on the upper or the lower part immediately Effects the whole object and the thinner it is the faster the heat actually is Propagated I mean loss of the heat is actually propagated inside and I don't really have to take into account the conductivity within this particular Object again simplification from the real life if I will have let's say a cube Then I will have to take into account that whatever my heat is dissipated From the outside it will take some time for the entire object to cool down. So again, no such complication my Square is infinitely thin, which means my heat is immediately effecting the temperature from the outside of this object immediately Conducted through the whole object. That's why it is thin Okay, that's good. Now. What else do I need? I Have dimensions. All right, I think it's time. We can actually start Calculating oh, yes, I do need something more. I need specific specific Heat Capacity of the object which is C because obviously if this is let's say piece of metal Then it will cool down during one particular Time and According to one particular law, but if it's a piece of let's say wood Or plastic then the probably the things will be different Okay, so I need the specific heat capacity now the way how I approach this Problem is the following during certain Amount of time Delta T I will have certain amount of heat Based on this temperature and this temperature temperature of the object and temperature of the outside Substance I will have certain amount of time of heat dissipated at the same time Since we have lost this amount of time of heat the temperature of the object must be lowered Now if I know by how much My temperature has changed I can calculate how much heat we have lost because I have this heat capacity and And the mass right so knowing mass and heat capacity and knowing the difference between two temperatures Before and after this period I can know how much I lost at the same time Since I know the temperature and outside temperature of object and temperature outside using the convection laws I Can calculate how much this substance actually took Heat and if I will basically just equate one to another I will have certain equation Based on which I can determine what is my function T right that that's a general general approach So amount of heat lost by the object should be equal to amount of heat consumed by convection in The substance okay, so let's calculate one and another and let's not pay attention to the sign because one probably is Positive another is negative doesn't really matter right now because when we equate them That would be or I mean we can all we can always say that okay They are positive and negative, but their sum is equal to zero same thing. All right doesn't really matter So I assume that this object is hotter and the substance is cooler So my heat goes from the object to the outside Substance, okay, so first of all first of all I will use the difference between these two temperatures and the law of convection To calculate how much heat we actually consumed by outside Substance and for this I need H which is this substance How is it called? I think I have the name convection heat transfer coefficient Convection heat transfer coefficient because this is needed for basically calculating how much heat will be Consumed by the substance through the convection Now as you remember from one of the previous lectures the amount of heat per Unit of time per unit of area. So the heat transfer Rage rate if you wish depends on this Convection heat transfer coefficient Convection heat transfer coefficient and the difference between the temperature So the temperature of the object is this temperature of substance is this All right That's what happens now if I have this rate of Heat transfer using the conductivity, but my object has certain area through which this heat is transferred because this is per unit of time per unit of Area, so I have to multiply it by area which is L squared times two right to L squared That would be the total amount of heat which Substance around this object is consuming using the convection All right provided this Coefficient of Heat transfer through the convection So this is my amount of heat which I have Consumed by the substance around This object Now let's think about what happens During the time Whatever was happening now. This is the rate. So I was talking about let's have certain amount of time delta t and To calculate how much we have Consumed during this period of time. I just have to multiply it by delta t so this is amount of heat energy is consumed by My outside world outside substance During the time delta t from the surface which has area to L square Top and bottom to L square, right? We're assuming this is a square and very thin square Now this is the temperature Now if my delta t is really small I can assume that the temperature is Constant during this period of time. I mean that's a general assumption in all the calculus So we are assuming that that the delta t or Increment of the argument is relatively small. So the function doesn't really change much during this increment And then we will go to a limit When when delta t goes to zero and then it will be the regular kind of an assumption Now let's calculate how much We have well we calculated how much we lost but now During the same amount of time if we know that we have lost temperature from t t to t of t plus delta t We are going to differential equation equations as you understand Now so during this period of time the function This function gives me change of the temperature Now if temperature has changed in the object itself Considering I know the Specific heat capacity and mass you know that the amount of Heat which I have lost would be t of t plus delta t minus minus t of t that's the difference of Temperature this is how much we have lost now if we will multiply it by specific heat capacity and mass That would be amount of heat which we have lost right? to increase if you remember to increase the temperature by a unit of temperate by by a unit of temperature we need amount of Object the mass of the object and specific heat capacity That basically definition of specific heat capacity so these two are supposed to be equal That's what the most important part of it Okay, from which everything follows very very easily if you divide this equation by delta t you will have C times m times t of t plus delta t minus t of t divided by delta t equals to to l square h T of t minus t s t s is a constant. That's the substance around the object temperature and Obviously you understand we go to a limit with delta t goes to zero in which case this is basically a definition of the derivative t of t By the way differential equations derivatives whatever whatever the calculus actually has in its Chest of tools These are the major tools used in physics everywhere. You will see differentials Derivatives differential equations etc etc the whole physics is based on this well the whole Apparatus of calculus was invented for the purposes of the physics by Newton and Leibniz and followers and some people even say that the whole mathematics is just Part of the physics, but okay, let's not go into this But in any case this is obviously a nice and very simple differential equation Which I'm going to solve right now and you will see where the exponential comes from First of all, let me simplify my my work I don't like all these are all constant. This is the constant which is the Specific heat capacity of the object whatever it's made of This is the mass. It's also given this is Convection heat transfer coefficient of the surrounding Substance and this is the sub the dimension of the of the object. It's a square L by L, right? So let me just put all of them together So I will put something like 2l square h divided by c times m Let's call it alpha whatever Now next thing I will simplify it x of t is equal to t of t minus ts this thing It's simpler if I will put it x of t why? because dx of t By dt is equal to dt of t by dt Because this is a constant, right? So subtract the constant from the function doesn't change its derivative. So This equation can be rewritten as now dt by dt is the same as dx by dt Equals to alpha times x of t Much simpler, right? And obviously we can solve it very easily Put dt over there and x t over here. So we have dx of t to x of t equals alpha t dt And now we can integrate it Now what is this? Well, this is derivative of logarithm, right? It's logarithm of x of t plus constant is equal to alpha t Now this is I don't want to use cc is already used somewhere Whatever the constant is All right So since it's just any kind of a free constant whatever it is We can put it here as well same thing plus minus doesn't matter. It's just any constant because we are Integrating, right? It's a indefinite integral All right, so we will define the coefficient beta a little bit later now obviously I have to Resolve it for x of t by doing what by raising e to this A logarithm of f of f of x and I will have f of f x of t is equal to e to the power of Alpha t Times again, this is any kind of a constant. So it's just another Coefficient times gamma right because we are raising to the power So that's why it's multiplication here all right now alpha we know what alpha actually is it's a What is it? It's 2l square h Divided by cm Okay, well actually I had to put the minus I was talking about minus and plus Sign so we basically understand that this is a negative thing because the temperature goes down, right? so The only thing is we don't know what gamma is right? Well, we can obviously Determine it by Substituting t is equal to zero when t is equal to zero at the beginning of the time This is one and the temperature x of zero is equal to x Zero I was talking about this in the very beginning. That's the beginning from which we start cooling our object, right? So that's basically what gamma is, right? I think I called it t zero not x zero All right, so basically we have solved everything and our answer to our problem is x of t well instead of x of t I will use now this t of t minus T s equals to this coefficient This well, that's not exactly true x of t is equal to all the T of zero minus t s, right? Which means t zero minus t s that's what gamma is So I have t zero minus t s times e to the power of Again, what is it minus to l square h divided by c m t t to the power So here is our exponent Now if you wish you can write it differently. We need Okay, that would be easier Again if t is equal to zero, this is one Which means my sum would be t s plus t zero minus t s so I will have t zero so a time t is equal to zero my temperature is equal to t zero Now if t is equal to infinity e to the power minus infinity is zero, right? Well, obviously we are talking about limits, right when t is Indefinitely increasing now this is going to zero, which means my temperature Eventually as the t is increasing will be closer and closer to this this component will be closer to zero So my temperature will be close and closer to temperature of the surrounding area but in any Finite moment of time it will not be equal so we need infinity actually so in the infinity My cattle will cool down to exactly the temperature of the surrounding air But obviously after Significant amount of time the temperature of the cattle will be insignificantly Greater than the temperature of outside air and we can actually complete the disregard this difference but anyway, this is the confirmation of our original logical kind of thinking that the graph is supposed to be Something like this, but this is T s and this is T Zero the temperature will go down Asymptotically approaching the temperature of the surrounding substance Okay, that's it Obviously we can put maybe I will put it in in the textual part of this lecture some calculation for a concrete case But in any case this is a very important Problem in that respect that it's using the laws which we already know which means the law of conductivity and the law of convection and the heat capacity and We're using a good mathematical apparatus of differential equations So just one more confirmation to whatever I was talking called the time that on the same unizord.com website There is a mass proteins course, which is a prerequisite. The mass is a prerequisite for the physics course. So Make sure you know everything whatever isn't that course. All right, that's it for today. Thank you very much and good luck