 Now, before Stephens Boltzmann came up with this expression, Newton experimentally found out something which was very useful earlier. What Newton found out was the rate of cooling of an object is proportional to the difference in the temperature between body and surrounding. Basically, how fast something is cooling down is directly proportional to what is the temperature difference. So, what Newton has told us that dq by dt, the rate at which something is cooling down is proportional to temperature of body minus temperature of surrounding. If I put a proportionality constant, then dq by dt should be equal to, all of you able to understand, right? Any doubts? I have a doubt. Quickly ask me. Yes, sir. If the temperature of the body is equal to that of the surroundings, will heat transfer still take place equally or will it stop? Heat transfer, it will be equilibrium. The amount of heat something is releasing will be equal to the amount of heat something is absorbing. You are not able to notice. So, it will be happening equally in both directions. Yes. Have you learned chemical equilibrium, chemistry? Yes, sir. Something like that. This is called physical equilibrium. Yes, sir. Thank you, sir. So, dq by dt will be some constant times tb minus ts. Okay. Now, we have learned in, let me go to the earlier slide. But before I go to, please write this down, the rate at which the object is cooling should be equal to c times tb minus ts. This is what Newton has told us before Stephens came up with the equation. This is called basically Newton's law of cooling. Okay. Fine. Now, let me go to the earlier slide and show you what is the problem here. So, here is the earlier slide. Here, dq by dt is equal to, this is a constant sigma into e into a is equal to constant times t to the power 4 minus ts raise to power 4. Okay. So, what Stephen has told us is this. Stephen has told us dq by dt is equal to some constant, let's say this is c1 and this is c2, t raise to power 4, tb raise to power 4 minus ts raise to power 4, okay. And this is actually a law, what Newton has found out, this is an experimental stuff. He has plotted graph and found out this. And this is, this came with some logic. What I'm saying is both the equations are correct, but there is an assumption in Newton's law of cooling. Please write down the assumption is temperature of body and temperature of surrounding are close to each other. When I'm saying close each other doesn't mean that they are equal. So temperature of body could be maybe just 80 to 100 degree Celsius away from temperature of surrounding, but it cannot be like 100 degrees. It cannot be, let's say 300 degree Celsius away from the surrounding or 1000 degree Celsius away from the surrounding temperature of body as long as it is between, you can set a 5 degree Celsius to 100 degree Celsius. If that is a temperature difference between body and surrounding, you can use Newton's law of cooling. Now, how can I get the Newton laws cooling from Stephen Boltzmann law? So what you can do is, you can rearrange the term, you can write it as C2, this is a square minus b square. Okay, so I can write this down as Tb raised to power 2 plus Ts raised to power 2 Tb2 minus Ts2. Then I can simplify it as Tb square plus Ts square Tb plus Ts into Tb minus Ts, okay. This I'm writing as dQ by dt. Now what approximation you have to make here is that if Tb is close to Ts, by the way tell me one thing, is Tb constant with time or Tb changing with time? Tell me quickly, is Tb, temperature of the body, changes with time or it is a constant? It changes, right? So surrounding temperature you can say it is constant, surrounding is big, you can't change the temperature of the surrounding, okay. Slowly you can't change it, of course greenhouse effect and everything slowly changes it, but you can't change it suddenly. So temperature of body keeps on changing, but then if temperature of body remains close to the temperature of surrounding, I can say that Tb square plus Ts square is roughly equal to 2 times of temperature of surrounding square and Tb plus Ts, I can say it is roughly equal to 2 times of surrounding temperature. So all the Tb is changing, but the change is not relevant because it is very close to the surrounding temperature, so I'm approximating this as a constant, 2 times the surrounding temperature square and that as another constant, 2 times Ts, okay. So you can modify everything and you can arrive at Newton's law of cooling, dQ by dt will be equal to another constant, this is a constant, this is also a constant and this was constant anyways. So combining 1, 2, 3 constant, let me call it as C1, Tb minus Ts, okay. This is the way you can arrive at Newton's law of cooling from the Stephen Boltzmann law, okay. Now how it can be used is this. If dQ by dt is C1 times Tb minus Ts, I can write down dQ as ms dt, dtb, sorry. Now tell me, I'm talking about cooling here, okay. This is Newton's law of cooling, so I'm talking about cooling. So dtb is less than 0 or it is greater than 0? Changing temperature is less than 0 or more than 0? Less than 0 because final temperature is less than the initial temperature, so final minus initial has to be less than 0, dt or delta of anything is final minus initial. So if it is cooling down, final temperature is less than the initial temperature. So final minus initial has to be less than 0 since Tf is less than Ti, okay. So that is why dQ has to be minus of that because right hand side is a negative quantity, left hand side is a positive quantity. So dQ by dt is minus of ms dtb by dt, like that, okay. Now I can substitute this value over here. By the way, why do you think I'm substituting dQ by dt like this? Because right now there are how many variables, three variables are there, q is there, t is there and tb is there, all three are changing. I want to decrease the number of variables, that is why I'm writing dQ by dt in terms of already which is a variable tb and t, okay. So when I write like this, I'm reducing the number of variable. So if I reduce the number of variable, the solving of equation becomes simpler. You will see how, okay. So now I will rearrange the terms and I will, what I'll say that, I'll put this thing over here, m into s, of course there is a minus sign, I'm putting in denominator, there's a minus sign over here. I'm calling c1 by m into s as another constant because mass is a constant, specific heat is also constant. So why do I again and again write constant? So I'll keep that as another constant. So dt by dt is let's say another constant minus of c2 times t minus ts. So I have written t instead of tb, okay. So it is understood whenever I write t, I'm talking about temperature of the body, okay. Temperature of the body, which is at the surface is equal to temperature of the entire body. Otherwise, you can't write ms dt anyways will not get into so much finer details that you will understand when you solve numerical only, but right now a basic understanding is fine. So I can rearrange the term dt divided by t minus ts will be equal to minus of c2 times dt. So I can create a differential equation and solve it. If I write dq by dt as ms dt, that is why I'm, I have written like that. So my values will be from 0 to t and let's say initial temperature is t1 and final temperature is let's say t2 at time t, do you know what is the integral of dt by t minus ts? What is this integral? I said tell me what is the integral of dt by t? What is this? Okay. Please remember this. The integral of dt by t is ln t. Similarly, integral of dt by t minus ts is ln of t minus ts. Go from t1 to t2. This is equal to minus of c2t, okay. So I can substitute the limits. This Newton laws of cooling is very mathematical. So probably you may have to memorize few things if you're not very comfortable with the differential equations. But soon you'll be getting used to these kinds of differential equations. So ln A minus ln B is ln of A by B. This is equal to minus c2t. So ln is log to the base E, okay. So I can further simplify this as t2 minus t1 divided by t2 minus ts divided by t2 minus t1 is equal to e to the power minus c2t. So I can say that t2 minus ts is t2 minus t1 e to the power minus c2t or t2 is equal to ts plus t2 minus t1 e to the power minus c2t. Is this derivation done in your school? Yes or no? No, sir. No, sir. Okay. So probably for your, I mean, but then this is there in your, let's say in your textbook, okay. So that is why I have done it. So probably in your school UT, it may not be asked, okay. But then if you look back, let me scroll through, look back from where we have started, we have started from Newton's law of cooling, dq by dt is a constant times difference in temperature between body and surrounding. Then I am saying that if dq is the amount of heat, dq should be equal to minus ms dt. So dq by dt will be equal to minus ms dt by dt. So I have substituted that and why I have substituted that so that I can reduce the number of variables. Once I have substituted that, it is now flowing like, you know, flowing with the logic. I am just going with what I should be doing. So here I am separating the variables. I am putting all the t terms left hand side and the time terms on the right hand side and then I am integrating and I will just keep on doing that finally I am getting this, okay. So that's how it is. And you can see that when t tends to infinity, e to the power minus e to t tends to zero because e to the power minus infinity is zero. So if this term is zero, t2 will be equal to surrounding temperature only. So you can see that after a long time, temperature of the body becomes equal to the surrounding temperature, which is what we have expected. And when t equal to zero, e to the power minus e to t is equal to one. So you can see that t2 was equal to, just once again, what is t1, yes, when t equal to zero, you can substitute and see what is the value of t2. Anyways, let us move to the, people are still joining is, Ronan is joining again and again for some reason, he's leaving then joining, leaving and running. Why? Could be internet problem. Is it reliance? All right. So this is how you have to do this. Is there any doubts? Anything? Sir, shouldn't it be ln t2 minus ts or ln t1 minus ts? Yeah, that is what I was wondering somewhere I made a silly error. So this is t2 minus t1. And this is t1 minus ts. Good that you helped me find that out. So this is t1 minus ts. So this is t1 minus ts, let us do that correction, this is t1 minus ts, so t1 minus ts. So now you can see that when you put t equal to zero and e to the power minus t2 is one, it becomes t2 is equal to ts plus t1 minus ts. So I'm getting t1 only. This is what we have assumed at t equal to zero, it's temperature is t1. Now in our school curriculum, this kind of differential equations solving and everything is not expected. But then there is an approximation for Newton's law of cooling. Based on that approximation, a lot of numericals are getting asked. So this is one of the favorite questions in your school exams. What they do is that rather than integrating it, they create a linear equation out of it. For example, what I'm trying to say is, all of you please write down dt by dt is equal to minus of constant times t minus ts. So in our school textbook, there is an approximation, please write down, approximation to above differential equation or you can say approximation to the Newton's law of cooling. So dt can be written as delta t divided by whatever time it has taken is equal to minus of c. What can you write for capital T? T is the temperature of the body which keeps on changing. So what should I write t as? Initial temperature or the final temperature? Final temperature. Okay, so it could be initial, it could be final or it could be in between. Could be anything. So I am playing safe. I am saying t is equal to the average temperature, t1 plus t2 by 2 minus ts. Okay. So you don't need to solve this like a differential equation. You can use this linear equation to solve numerical. Please write down, this is the most important part of your, at least a school exam. One numerical on this is expected. This and derivation of Carnot cycle efficiency. So I'll talk about that also maybe in the next class, but right now focus on this. Any doubt on the approximate formula for the Newton's law of cooling? Anyone? You can click yes or no. Okay. So with respect to theory, that's it. But we will end the session with two numericals on this approximate formula since this is very important. So I'll just project a couple of questions. So let us solve them. Is the blackbody radiation done? Is the, what do you call? Is the blackbody radiation done that lambda into t is constant that equation is done in your school? Anyone can answer? Sir, they finish blackbody radiation. Jesus explained in brief in our class. Have they talked about lambda into t is a constant? No, sir. Anyways, so some portion is left over. Anyways, you can see there is a question in front of your screen. Please solve it. If you get it right the first time, you don't need to do Newton's law of cooling again. And the formula which you have to use is in front of you. This formula. So will we have to take the temperature in Kelvin? I have to take temperature in Kelvin. No, you don't need to. If you take all temperature in degrees Celsius because there is a difference in temperature everywhere. The right hand side is also difference between this temperature, average temperature and T s. Right, left hand side is also delta T. So you don't need to convert in Kelvin. So what is C? What is? C value. C is not given to you. That is why there are two scenarios given in the problem. So you have to get rid of C. Two equations, two variables you will get. C is a variable when you write down the two equations. So is it 0.7 minutes? One more minute guys. I'm waiting for everyone to solve it. So you're getting 42 seconds. 42 seconds. Okay. 42 seconds. Yeah, 42 seconds. So let me write down the equation. Please, all of you pay attention here. There are two scenarios that are given in the problem. Isn't it? The pan filled with hot food cools from 94 to 86 in two minutes. This scenario one, when the room temperature is 20. How long will it take to cool down from 71 to 69? So 94 to 86 is scenario one and scenario two is 71 to 69. So without wondering about how to solve the equation, the first step is to write down the equation. Okay. So the equation for the case one. What is case one? 94 to 86. So 94 to 86 delta T is what? Delta T is final temperature minus initial temperature. So 86 minus 94 divided by delta T, which is the time it has taken. So I can write down as two into 60 seconds. You need not convert in second also, but just to place if you're converting in seconds. And that's a good practice. Minus C times T1 plus T2 by two average temperature, which is 94 plus 86 by two minus temperature of surrounding. Surrounding is your room temperature minus 20. So this is your case number one equation. Case number two is 71 to 80, 71 to 69. So 71 minus 69 divided by, now time is not given. In fact, they're asking what is time. Well, keep it as T. Now this constant does not change with respect to which scenario I'm talking about. So constant remains C only. Okay. That into now T1 is, sorry, here also I should write 69 minus 71. It is always final minus initial. Now here T1 is 71, T2 is 69. So I'll take average of these two temperatures now. There are two different scenarios. Mine is the room temperature remains 20 for both the scenario. Now to get rid of C, what I can do is that I can divide left hand side and the right hand side. So when I divide C disappears. And then you can get the value of T. Okay. So even if you write very neatly these two equations, you'll get most of the marks from the numerical. I hope all of you are able to solve and get the answer. The answer is roughly 42 seconds. There's one more numerical. Very similar to this. So that those who did not get the first numerical, you may try to get this right now so that feel confident. This one, attempt this, all of you. Can you check the final answer from your school textbook? This is the last question of thermal properties of matter. Let me know what is the answer given in your textbook. Sir, I'm getting 660 seconds. Yeah, can you check the final answer from your book? I don't have access to the final answer. How much it is? 9 minutes. 9 minutes is the answer given? Yes, sir. 9 minutes. Yeah, so 9 minutes is 540 seconds. 540 seconds, yes sir. Okay, so what happens is that they don't expect you to convert minutes into seconds. Since time is given in minutes, so they are probably expecting you to get the final answer in minutes only. But then we will follow our usual process. We will get the time in seconds and then you can convert back into minutes. Just to avoid any silly errors, we will follow the SI system. So I'll quickly write down the equation. DT by DT is equal to minus of C times T minus TS. The approximate formula for this is delta T by this minus C times T1 plus T2 by 2 minus TS. So there are two cases, case number one, 80 to 50. So 50 minus 80 divided by five minutes is minus of C times 50 plus 80 by 2. Temperature of surrounding is 20 and then it is 60 to 30. And this time I have to find what is T minus C, 30 plus 60 by 2 minus 20. So again you divide it, C and C will disappear. And just rearrange the terms, you'll get the answer. Okay, very simple and straightforward. So these kind of numericals can be asked, okay? All right friends, so that's it for today. Thank you sir. Thank you sir. Thank you sir. Thank you sir.