Newton's Law of Cooling

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Take a hot body, say a cup of tea, at eighty degrees Celsius, in a room at twenty degrees. In ten minutes, it cools to sixty. Now, how long will it take to cool from sixty to forty? Another twenty degrees, so another ten minutes, right? Hmm, not so fast. Today: Newton's law of cooling, and by the end you will know the real answer. Cooling questions come up often in JEE, so let's begin.

0:34Newton's law: the rate depends on the excess

So, a body at temperature capital T sits in surroundings at T naught, the T with a small zero below it. The difference, T minus T naught, is called the excess temperature, okay? Today is all about this excess. Now think. The same body, once at sixty degrees and once at thirty degrees, in a room at twenty. In which case does its temperature fall faster? At sixty! Yes, exactly! The bigger the excess, the faster it cools. That is Newton's law of cooling: d T by d t equals minus k, times T minus T naught. d T by d t is the rate of change of temperature, the minus sign says the temperature is falling, and k is a positive constant, the cooling constant. And see the condition in brackets: it holds only when the excess is small. Clear? What decides k? k equals h times capital A, over m times s. Capital A is the surface area, m is the mass and s is the specific heat, so m times s is the heat capacity. And h depends on the nature of the surface. So more surface means faster cooling, and more heat capacity means slower cooling. Got it?

1:49Exact form: equal factors in equal times

Now, solve that rate equation and you get the exact form. T minus T naught equals the bracket T i minus T naught, times e to the power minus k t. T i, with the small i below, is the initial temperature. So the excess dies away exponentially, okay? Now tell me. In every ten minutes, does the excess fall by the same number of degrees? Or by something else? Not by the same degrees, but by the same factor! Correct! Equal times multiply the excess by the same number. So the excess halves every log two over k. Here l n on screen is the natural log, and we read it as log. Remember this halving idea. Right? Take the log of both sides, and log of T minus T naught falls linearly with time, t. So its graph against t is a straight line, with slope minus k. With me?

2:43Average form, for short intervals

The exact form needs logs. For short intervals there is a quicker average form. The body cools from T one to T two, the small one and two below T, in time t. Then T one minus T two, over t, equals k times, the bracket: T one plus T two over two, minus T naught. Okay? The left side is the average rate of fall. The right side is k times the average excess: the average temperature minus the surroundings. Same law, with averages. Clear? Now, a question. Can we use this average form over any interval, even a very long one? No, no, careful! Over a long interval the rate changes a lot, so the average is only an approximation. Use it for short intervals; for long ones, use the exact form. Got it?

3:36Where it comes from: Stefan's law

So where does this law come from? One neat way to see it is through radiation. By Stefan's law, the net rate of heat loss to the surroundings is e sigma capital A, times T to the power four minus T naught to the power four. And here the temperatures must be in kelvin, okay? Small e is the emissivity, equal to one for a black body. And sigma, the Greek letter, is here the Stefan-Boltzmann constant. Got it? Now think. If the body is only slightly hotter than the surroundings, what does T to the power four minus T naught to the power four become? It becomes about four T naught cubed, times delta T, okay? Delta T, with the little triangle, the Greek capital delta, is the small excess, T minus T naught. And the wavy equals sign means approximately. Right? So the net rate of loss is about four e sigma capital A, T naught cubed, times delta T, okay? The sign that looks like an open eight means proportional to. So the rate of loss is proportional to delta T, and that is Newton's law! This is exactly why the law needs a small temperature difference. Clear?

4:50Example 1: time to cool from 90 to 45 C in a 30 C room

Now, a course example. The small circle and C mean degrees Celsius. Read it with me: a body at ninety degrees, in a room at thirty degrees, reaches sixty degrees in eight minutes. When does it reach forty five degrees? Use the exact law. Pause the video, and try it. Hint: track the excess. Done? Okay, let's check. At the start, the excess is ninety minus thirty, sixty. After eight minutes, it is sixty minus thirty, thirty. The arrow means goes to. So the excess halved in eight minutes. Okay? Now, forty five degrees. The first thirty degrees took eight minutes, and now only fifteen more degrees to fall. So just four more minutes, twelve in total, right? No, no, careful! Cooling is not steady. Forty five degrees means an excess of forty five minus thirty, fifteen. From thirty to fifteen, the excess halves again. With me? And equal factors take equal times. So another eight minutes: eight plus eight, sixteen minutes from the start. See? The halving idea made it a one-liner!

6:05Example 2: 60 to 40 C takes 17 min, not 10, both forms

Remember our puzzle? Let's finish it. The body cools from eighty to sixty degrees in ten minutes, in a room at twenty. How long does it take to cool from sixty to forty? Use the average form, then the exact form. Pause the video, and try it. Done? Okay, let's check. Average form first, on the first interval. The fall is twenty degrees in ten minutes, so twenty over ten. The average temperature is seventy, so the average excess is seventy minus twenty. So twenty over ten equals k times, seventy minus twenty. The double arrow means so: k comes out as one over twenty five. Right? Now the second interval: again a fall of twenty degrees, in time t. The average temperature is fifty, so the average excess is fifty minus twenty, thirty. Twenty over t equals one over twenty five, times thirty, so t is about sixteen point seven minutes. Not ten at all! Now the exact form. From eighty to sixty degrees, the excess went from sixty to forty, in ten minutes. That is a factor of one point five, okay? From sixty to forty degrees, the excess goes from forty to twenty: a factor of two. Equal factors take equal times, so we compare the logs. The time is ten times log two over log one point five: about seventeen point one minutes. Okay? So the answer is about seventeen minutes, not ten. That was the trap in our puzzle: as the excess shrinks, the cooling slows down. And both forms agree closely: sixteen point seven and seventeen point one. Clear?

7:51Example 2 on the cooling curve

Now let us see this on a graph. This is the same body, cooling from eighty degrees in a room at twenty. Temperature in degrees Celsius goes up, and time t in minutes goes along. The dashed line is the room temperature, T naught, twenty degrees. With me? The first dot is the start: eighty degrees at time zero. The next dots sit at ten minutes, at sixty degrees, and at twenty seven point one minutes, at forty degrees. Okay? So twenty seven point one minus ten is seventeen point one minutes. Exactly our answer from the exact form. Perfect! And look at the shape. As T comes close to T naught, the curve flattens: the excess shrinks, so the cooling slows down. Clear?

8:37Quick revision: rate law, exact form and average form

So, quick revision. For a small excess, d T by d t equals minus k, times T minus T naught. In the exact form, the excess falls by equal factors in equal times, not equal degrees. And for short intervals, use the average form, with the average excess. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!