If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Newton's Law of Cooling

Newton's law of cooling can be modeled with the general equation dT/dt=-k(T-Tₐ), whose solutions are T=Ce⁻ᵏᵗ+Tₐ (for cooling) and T=Tₐ-Ce⁻ᵏᵗ (for heating).

Want to join the conversation?

  • aqualine seed style avatar for user Karsh Patel
    Does that mean that ice cream pulled out from a refrigerator at -4 C' will get hotter more quickly than that pulled out from a refrigerator at 0 C'?
    (17 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Yamanqui García Rosales
      Yes, since the temperature difference will be greater with the cooler ice cream, that one will be subjected to a faster increase in temperature.

      Still, by the time it gets to 0℃, the rate of temperature increase will be the same as the ice cream that was originally at 0℃, so the colder one will always take more time than the not so cold to reach the same temperature.
      (30 votes)
  • female robot amelia style avatar for user where
    dT/dt=-k(T-Ta) i don not understand the negetive k,can't it just be positive?
    because later we need to take the absolute value and write two functions according to the object is hotter or cooler?thanks in advance!
    (6 votes)
    Default Khan Academy avatar avatar for user
    • mr pants teal style avatar for user James Mick
      For Newton's law of cooling you do not need to have the negative sign on the k, but you do need to know/understand that k will be a negative number if an object is cooling and a positive number if the object is being heated. This makes intuitive sense as you would need a positive exponent to increase temperature and a negative exponent to decrease temperature. The main reason I can see for putting the negative k in is to keep you from forgetting it later.
      As far as the two equations go, I can tell you that I was able to solve a few problems using either equation. It requires a little bit of manipulation and you really have to think about what you are doing in order to achieve this, but it can be done. Know that if you perform it with the wrong equation, then you will end up with a negative t, which just means that you were going back in time to warm or cool your object. It is probably best to know that there are two equations, and when to use them in order to save yourself the mental anguish of having to perform these manipulations.
      To summarize, the negative sign is put in front of the k as a means to prevent you from accidentally omitting it later, and the 2 equations are to keep you from having to wrestle with even more awkward equations and ending up with a negative time.
      (17 votes)
  • leaf orange style avatar for user Techless
    At Sal starts to integrate, why do the dT and dt terms vanish in the process?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user farazbinferoz
    Doesn't the cooling depend on the other factors as well like the nature of matter?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • mr pants teal style avatar for user James Mick
      Absolutely, The k is a ratio that will vary for each problem based on the material, the initial temperature, and the ambient temperature. Most of the problems that I have seen for this involve solving for C, then solving for k, and finally finding the amount of time this specific object would take to cool from one temperature to the next.
      (10 votes)
  • mr pants pink style avatar for user xingxinyi
    What are the factors that influence the speed of the temperature to get cool?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • purple pi teal style avatar for user zhaolongwang
    If the cooling of the coffee is affected by external factors, the calculation is still accurate
    (3 votes)
    Default Khan Academy avatar avatar for user
  • female robot amelia style avatar for user Deb Jarz
    Please, can you use actual NUMBERS in reference to the LETTERS. If I could see NUMBERS I might actually understand. Just letters is so confusing. If you have a link to another video that shows numbers, please post here. Thank you so much.
    (0 votes)
    Default Khan Academy avatar avatar for user
    • leafers seed style avatar for user Travis Bartholome
      Head on over to the next video, entitled "Worked example: Newton's law of cooling," and you'll see Sal work a problem like this with numbers.

      Just on a side note, though, I'd be remiss not to point out that the way Sal solves this, using arbitrary constants, is probably the way that makes things easiest in the long run. Sure, we could "remove" two of the constants here (k and T_a) by replacing them with numbers. Then you have a number to look at instead of a letter (although we can't get around adding the constant C to the mix). But ultimately, writing a letter is really no different conceptually than writing a number -- they're just different symbols for a constant.

      Also, defining the constants first is not particularly helpful if you're trying to solve an initial value problem or otherwise trying to fit your equation to real-world situations. Typically you'll have no idea what the constants are, but you'll know what values the function should have at different points along the t axis. You'll run into constants extremely frequently that are similar to the ones in this video. C is an integration constant, and k is a proportionality constant. Both show up in almost every exponential model you'll see in a differential equations course, and I'm not sure you can get by without knowing how to solve them this way.

      Hopefully all that doesn't sound rude -- I don't intend it to be. But being uncomfortable using letters/symbols instead of numbers will definitely hold you back in pretty much every branch of mathematics.
      (11 votes)
  • purple pi purple style avatar for user Comett8
    how and why would the equation be if the heat from the hot cup changed the temperature in the room?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Yamanqui García Rosales
      It would be a completely different, and much more complicated equation. As you already noticed, one of the simplification that Newton's Law of Cooling assumes is that the ambient temperature is constant, but it's not the only simplification. Newton's Law of Cooling also assumes that the temperature of whatever is being heated/cooled is constant regardless of volume or geometry.

      If you wanted to create a more realistic (and therefore more complicated) model of temperature exchange, the Diffusion Equation is probably a good starting point, since it does considers geometry.
      (4 votes)
  • blobby green style avatar for user pranav.mayelam
    At we can see the finished formula for when the temperature of the object is greater than our ambient temperature. I still don't understand what all the constants mean. What does each constant in the equation refer to?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • aqualine tree style avatar for user Xristina Nolita
    If T=Ta then we have T-Ta=0 so we can't write ln(T-Ta) or 1/T-Ta. What do we do then? My guess is to start solving the equation saying that T is not Ta because in that case dT/dt would be 0.
    (2 votes)
    Default Khan Academy avatar avatar for user

Video transcript

- [Voiceover] Let's think about another scenario that we can model with the differential equations. This is a scenario where we take an object that is hotter or cooler than the ambient room temperature, and we want to model how fast it cools or heats up. And the way that we'll think about it is the way that Newton thought about it. And it is described as Newton's Law of Cooling. And in a lot of ways, it's common sense. It states that the rate of change of temperature should be proportional to the difference between the temperature of the object and the ambient temperature. So let me write that in mathematical terms. So Newton's Law of Cooling tells us, that the rate of change of temperature, I'll use that with a capital T, with respect to time, lower case t, should be proportional to the difference between the temperature of the object and the ambient temperature. So that is a mathematical description of it. And once again, it's common sense. If something is much, much hotter than the ambient temperature, the rate of change should be pretty steep, it should be declining in temperature quickly. If something is much, much cooler, it should be increasing in temperature quickly. And if something is close, if these two things are pretty close, well maybe this rate of change shouldn't be so big. Now I know one thing that you're thinking. You're like, okay, if the temperature is hotter than the ambient temperature, then I should be cooling. My temperature should be decreasing. And a decreasing temperature would imply a negative instantaneous change. So how will this be a negative value in the case where our temperature of our object is greater than our ambient temperature? And the way that that would happen is, you would have to have a negative k. If you don't like thinking in terms of a negative k, you can just put a negative right over here and now you would have a positive k. Now it makes sense. If our thing is hotter, if it has a higher temperature than the ambient temperature, so this is a positive, then our rate of change will be negative, will be getting cooler. If it was the other way around, if our temperature of our object is cooler than our ambient temperature, then this thing is going to be a negative, and then the negative of that is going to be a positive, we're assuming a positive k, and our temperature will be increasing. So hopefully, this makes some intuitive sense. And our constant k could depend on the specific heat of the object, how much surface area is exposed to it, or whatever else. But now I'm given this, let's see if we can solve this differential equation for a general solution. And I encourage you to pause this video and do that, and I will give you a clue. This is a separable differential equation. So I assume you've had a go at it, so let's now work through it together. So, we just have to algebraically manipulate this so all my Ts and dTs are on one side. I should say, so all my capital Ts and dTs are on one side, this is going to be a little bit more confusing because I have a capital T and a lower case t. Capital T for temperature, lower case t for time. But hopefully we'll be able to work through it. And then I'm going to have all my time differentials and time variables on the other side. So one thing I could is I could divide both sides by T minus ambient temperature, minus T sub a. Remember this is just going to be a constant based on what our ambient temperature is. We're going to assume our ambient temperature doesn't change as a function of time, it's just such a big room that our cup of tea is not going to actually warm up the room. So that's just one of these assumptions that we're going to make. So if we do that, if we divide both sides by this, we are going to have... So I'm going to divide both sides, I'm going to do this in a new color. If I divide both sides by that, I get one over T minus T sub a, and let me multiply both sides times the time differential. So I'm going to have, that dT, our temperature differential. Times our temperature differential, is going to be equal to negative k times our time differential. So once again, to separate the variables, all I did was divide both sides by this, and multiply both sides by that. Now I can integrate both sides, we've seen this show before. So I can integrate both sides. And the integral of this is going to be the natural log of the absolute value of what we have in the denominator. And you can do u substitution if you want. If we said u is equal to T minus T sub a, then du is just going to be one dT, and so this is essentially, you could say the integral of one over u du, and so it would be the natural log of the absolute value of u, and this right over here is u. So this is the natural log of the absolute value of T minus T sub a, is equal to, and once again I could put a constant here, but I'm going to end up with a constant on the right hand side too so I'm just going to merge them into the constant on the right hand side. So that is going to be equal to, now here, this is going to be negative kt, and once again we have plus C. And now we can raise e to both of these powers, or another way of interpreting this is if e to this thing is going to be the same as that. So we can write this as, the absolute value, let me do that in that same blue color. We can write this as the absolute value of T minus T sub a is equal to e, something about e I always think of the color green. e to the negative kt plus C. This of course is the same thing as, this is equal to e to the negative kt, we've done this multiple times before. Negative kt times e to the C power. And we could just call this another arbitrary constant. If we called this C1, then we could just call this whole thing C. So this we could say is Ce to the negative kt. So at least it's starting to resemble what we did when we were modelling population. We'll see it's a little bit different. Instead of just temperature on this left hand side, we have temperature minus our ambient temperature. And so, we can do a couple of things. If, in a world, say we were dealing with a hot cup of tea, something that's hotter than the ambient temperature. So we could imagine a world where T is greater than or equal to our ambient temperature. So that means this is hot, or it's hotter, I guess we could say. So if we're dealing with something hotter than the ambient temperature, then this absolute value is going to be positive or the thing inside the absolute value is going to be positive. So we don't need the absolute value. Or the absolute value of it is going to be the same thing as it. And then we can just add T sub a to both sides, and then we would have our temperature, and I can even write this as a function of time, is going to be equal to this business, is going to be equal to Ce, let me do that in that same color. Ce to the negative kt plus T sub a. All I did is I'm assuming that this inside the absolute value is going to be positive, so the absolute value is not going to change the value. And I added T sub a to both sides to get this. So this right over here is going to be our general solution, in the case where we start with something that is hotter than the ambient room temperature. And if we want to look at the case where something is cooler than the ambient room temperature, so that's the situation, let's say T is less than our ambient room temperature. Then the absolute value of T, then this thing over here is going to be negative, and so the absolute value of it's going to be the negative of that. So then this up here results in T sub a minus T, that's going to be the same thing as the absolute value, it's going to be the negative of the negative. So then that is going to be equal to e to the negative k plus, actually let me just do it... T sub a minus T is going to be equal to Ce to the negative kt, so this is equal to that. I'm just assuming that T is less than T sub a. And so then, to solve for T, you could add T to both sides and subtract this from both sides. You would have T as a function of t is going to be equal to, let's see, if this went onto that side and this goes over here, you would have T sub a minus Ce to the negative kt. Did I do that right? So yep, that looks right. So this is the situation where you have something that is cooler than the ambient temperature. So this right over here, based on the logic of Newton's Law of Cooling, these are the general solutions to that differential equation. In the next video we can actually apply it to model how quickly something might cool or heat up.