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Current time:0:00Total duration:12:23

Video transcript

let's now actually apply Newton's law of cooling so just to remind ourselves if capital T is the temperature of something in Celsius degrees and lowercase T is time in minutes we can say that the rate of change the rate of change of our temperature with respect to time is going to be proportional and I'll write a negative K right over here if we assumed our constant K is positive then a negative K it's going to be proportional to the difference between the temperature of our thing and the ambient temperature in the room and once again why I'd have a negative there well because if the temperature of our thing is larger than the temperature of the room we would expect that we would be decreasing in temperature that we would have a negative rate of change that set temperature should be decreasing with time if on the other hand our temperature is lower than the ambient temperature of the room then this thing is going to be negative but we would want a positive rate of change of temperature things would be warming up and so that's why a negative of a negative would give you the positive so this right over here this differential equation we already saw it in the previous video on Newton's law of cooling and we even saw a general solution to that and the general solution that I care about because we're now going to deal with the scenario where we're putting something warm in a orce we're going to put a warm bowl of oatmeal in a in a room temperature room so given that we're going to assume the case that we saw in the last video where our temperature is greater than or equal to the ambient temperature and in that situation our general solution boiled down to it boiled down to temperature as a function of time is equal to some constant times e to the negative KT negative KT plus our ambient temperature plus our ambient temperature once again we figured this out in the last video now let's actually apply it so I said where we're dealing with the scenario where our temperature is greater than or equal to the ambient temperature so let's assume we're in a scenario let's assume a scenario where ambient temperature is 20 degrees Celsius so we assume that doesn't change that the room is just large enough that even if something is warmer is put into it that the ambient temperature does not change and let's say that the thing that we put in it our warm bowl of oatmeal let's say it starts off the moment we put in the room at time equals zero is 80 degrees Celsius and let's say we also know just from previous tests that after two minutes after two minutes it gets to sixty degrees Celsius so we also know that T of two is 60 degree Celsius and given all of this information right over here using Newton's law of cooling and using all of this information we know about how bowls of oatmeal that started at this temperature have cooled in the past we want to know how long will it take how many minutes will have to pass when you put an a degree bowl of oatmeal in the room how many minutes have to pass in order for it to get to 40 degrees using this model so let me write that down so how long how many minutes minutes four or let me say - cool - cool - cool - forty degrees Celsius and so I encourage you to pause the video now and try to figure it out so I'm assuming you have paused the video and you have had your go at it and the key is is to use all of this information right over here to solve for the constants C and K and then once you know that you essentially have described your model and then you can you can apply it to solve for the time that gets you to a temperature of 40 degrees Celsius so let's do that so the first thing we know is the ambient temperature is 20 degrees Celsius so this right over here this right over here is 20 degrees and so the most obvious thing to solve for or to apply is what what happens with T of zero and what's neat about T of zero when T equals zero this exponent is 0 e to the zero power is 1 and so T of zero is essentially going to simplify to C plus 20 degrees let me actually write down so T of zero which we already know is eighty degrees we already know is a degree Celsius and I'm just going to write eighty I'm just going to write eighty we'll assume it's in degrees Celsius that is going to be equal to that is going to be equal to when T equals zero this the e to the negatives or e to the zero is just going to be one so it's going to be equal to C plus C plus twenty and so if you want to solve for C you just subtract 20 from both sides of this of this equation and we are left with we are left with eighty minus 20 is 60 is equal to C 60 is equal to C so we were able to figure out C so let's figure out let's write what we know right now so we know that t let me do that in that magenta color we know that T of T and it's confusing upper case upper case T of lower case T temperature as a function of time is going to be equal to is going to be equal to them into that same color 60 e 60 e to the negative KT negative KT plus 20 plus our ambient temperature plus our ambient temperature now we need to solve for K and we could use this information right over here to solve for K T of two is equal to 60 degrees so if we make T is equal to two this thing is going to be sixty degrees so let me write that down so we could let me write it over here so have some space so we have 60 is equal to 60 is equal to 60 is equal to 60 e to the negative KT all this color switching takes time e to the negative KT plus au and we be careful that's a time two so that's e to the negative K times two that's a time equals two or I could write that e to the negative 2k e to the negative 2k and then of course we have our plus twenty and then we have our plus twenty and now we just have to solve for K again at any point if you feel inspired to do so I encourage you to try to solve it on your own all right so let's do this so if we subtract if we subtract 20 from both sides we get 40 is equal to is equal to 60 e e to the negative 2k negative 2k and DeVoe divide both sides by 60 you are left with you are left with two thirds 40 divided by 60 is 2/3 is equal to e to the negative 2k e to the negative 2k negative 2k all this color changing it takes work let me know if you all want me to keep changing I enjoy changing colors it just keeps it interesting on the screen but anyway so e to the negative 2k and actually let me scroll down a little bit so I have some more real estate to work with and now I could take let's see I could take the natural log of both sides and so I'll have the natural log natural log of 2/3 is equal to the natural log of e to the negative 2k is just going to be negative 2k that's the whole reason why I took the natural log of both sides and then to solve for K I divide both sides by negative 2 I get K is equal to negative 1/2 negative 1/2 natural log natural log of two thirds I just swap sides natural log of two thirds let me do that since I kept the colors going so long let me keep it that way natural log of two thirds and so we have solved for all of the constants so now we can rewrite this thing right over here we can rewrite it as so we decided from mini drumroll here we're not completely done yet we get T of T is equal to is equal to 6t e e to the negative K well negative K the negative and negative can be positive can be 1/2 natural log of two thirds so 1/2 natural log of two thirds negative K so negative of a negative so 1/2 natural log of two thirds which actually will be a negative value two thirds is less then II so you're going to have a natural log of it is going to be negative so it makes you feel good that the temperature is going to be is going to be going down over time and so plus au times T times T plus 20 plus 20 and so now all we have to do is we have to figure out what T gets us to a temperature of 40 degrees Celsius so let's solve for that so if we want this to be 40 for T is equal to it actually now I'm just going to stick to one color as we as we march through this part so for T is going to be equal to 60 e to the 1/2 natural log of two thirds t power plus 20 now let's see we can subtract 20 from both sides we get to 20 is equal to 60 I eat all that crazy business 1/2 natural log of two thirds times T now we can divide both sides by 60 and we get 1/3 20 divided by 60 is 1/3 is equal is equal to e to the one half natural log of two thirds times T now let's see we could take the natural log of both sides so we get the natural log of one-third is equal to 1/2 natural log of two thirds times T and then homestretch to solve for T you just divide both sides by one half natural log of two thirds so we get T is equal to this which is the natural log of 1/3 divided by one half natural log of two thirds well if you divide by one half that's the same thing as multiplying by two and then you're going to divide by natural log natural log of two thirds 2/3 now let's see if this actually makes a sensical a sensical answer so let me get going to calculator out actually I could just actually I could just use I could just use Google here so I had actually I already forgot I already forgot what it was natural one so I have natural log one-third over natural log of two thirds and the whole thing times two so I can type I can type too two times the natural log of natural log of one third natural log of one third divided by divided by the natural log of two thirds let's see what Google gets us all right it didn't what did I wear how did I mess up so this is equal to two times the natural log okay messed up the parentheses let me make this clear the natural log of 1/3 divided by the natural log of two thirds and then we're there you go if we were to round to the nearest hundredth there would be five point four two so five point four two minutes remember everything we were doing were in minutes so this right over here this is approximately equal to five point four to five point four two minutes and we are done that's how long it will take us to cool to 40 degrees