Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Okay, a puzzle to start. Picture a five hundred watt immersion heater, switched on in two kilograms of water. Five hundred joules every second, and yet the water gets no hotter; it stays at sixty degrees Celsius. So where is all that energy going? And when we switch it off, how fast does the water start to cool, with no area and no cooling constant given? This is set three, the hard set of mixed problems in this unit. Today, two of its heat problems, both built on one idea: the steady state. Questions like these come up often in JEE, so let's begin.

0:40Tool 1: heat, power and temperature change

So, tool one. Heat to change temperature: capital Q equals m s delta T, okay? Here m is the mass, s the specific heat, and delta T, with the Greek capital delta, the little triangle, is the change in temperature. Clear? For water, s is four thousand two hundred joules per kilogram per kelvin, written J k g to the power minus one, K to the power minus one. Right? Power, capital P, is heat per second. One watt, capital W, is one joule per second, written J s to the power minus one. Got it? Now, divide capital Q equals m s delta T by the time taken. What do you get? Heat per second on the left: that is power. And on the right, m s times d T by d t, the rate at which the temperature changes. So P equals m s, d T by d t. Okay?

1:39Tool 2: heat loss and the steady state

Look at this graph: a body cooling from eighty degrees Celsius in a room at twenty; the small circle and C mean degrees Celsius. Going up is capital T, the temperature; along the bottom is small t, the time in minutes. The dashed line is the surroundings, T naught, with a small zero below, and the orange dots are readings on the way down. With me? So tell me: where is this curve steepest, at the start, or near the end? At the start, yes, exactly! While the body is hot, it cools fast; near room temperature, the curve flattens. So, hotter body, faster heat loss; the double arrow means so. Clear? Newton's law of cooling makes this exact, for a small temperature difference only: d T by d t equals minus k, times T minus T naught, with k a constant for the body. The formula on the graph is its exact form: T minus T naught equals, T i minus T naught, times e to the power minus k t, where T i, with a small i below, is the starting temperature. Right? Now, back to our heater. It keeps pumping heat in, yet the temperature stays steady. So, heat in and heat out: what must be true? They are equal! That is the steady state: heater power equals the rate of heat loss. The energy leaks to the room as fast as it comes in. Got it? Now, the key idea for today. Switch the heater off. At that instant, the body is still at the same temperature, so at first it loses heat at the same rate. Only now, nothing comes in. Okay?

3:23Tool 3: radiation and the net loss

Now, tool three: radiation. The Stefan Boltzmann law: P equals e sigma capital A T to the power four. P is the power radiated, small e the emissivity, sigma, the Greek letter, the Stefan Boltzmann constant, capital A the surface area, and capital T the temperature, always in kelvin. With me? Here sigma is that constant, not surface tension: five point six seven times ten to the power minus eight watts per square metre per kelvin to the power four. And for a black body, the best emitter, e is one. Clear? But the body also absorbs radiation from its surroundings, at T naught. So the net loss is e sigma capital A, times, T to the power four minus T naught to the power four. Right? And radiation needs no medium at all, so it crosses even a vacuum. Got it?

4:19Example 1: initial cooling rate after a 500 W heater is off

Now, Example one: our puzzle! Here, a five hundred watt immersion heater keeps two kilograms of water steady at sixty degrees Celsius, in a room at twenty degrees. The heater is switched off. Find the initial rate of cooling. Pause the video, and try it first. Done? Okay, let's check. The water was steady, nothing changing. So just after switch-off, the rate of cooling is zero, right? No, no, careful! Steady means heat in equals heat out, not zero loss. The heater supplies five hundred watts, so the water at sixty degrees loses five hundred watts to the room. That is where the energy was going! With me? Just after switch-off, the water is still at sixty degrees, so it still loses five hundred watts. And now nothing comes in. Clear? So the rate of cooling, minus d T by d t, equals P over m s; the minus sign because the temperature is falling. That is five hundred, over two times four thousand two hundred. About zero point zero six zero kelvin per second; the wavy equals sign means approximately, and K s to the power minus one is kelvin per second. Okay? Times sixty: about three point six kelvin per minute. And a change of one kelvin is a change of one degree Celsius, so about three point six degrees Celsius per minute. No area, no cooling constant, not even the room's twenty degrees. Puzzle solved. Perfect!

5:56Example 2: power to keep a black sphere at 1000 K in vacuum

Now a harder one. Here is a black sphere of radius one centimetre, kept at one thousand kelvin by a heater inside it. It sits in an evacuated enclosure, air pumped out, at three hundred kelvin. Find the heater power. Pause and try it. Hint: steady state again. Done? Okay, let's check. There is no air around the sphere. So how can heat leave it at all? By radiation only, exactly! The vacuum stops conduction and convection. And the temperature is steady, so, just as in example one, heater power equals the net loss. Right? So P equals sigma times four pi r squared, times, T to the power four minus T naught to the power four; the dot means times. Four pi r squared, with pi the Greek letter, is the area of a sphere, and the sphere is black, so e is one. Okay? Now, one thousand minus three hundred is seven hundred. So just take seven hundred to the power four, right? Oh no, careful! Fourth powers first, then subtract. One thousand to the power four is ten to the power twelve, and three hundred to the power four is eight point one times ten to the power nine. Never seven hundred to the power four! Got it? Now the area. One centimetre is ten to the power minus two metres, so four pi r squared is four pi times ten to the power minus four: about one point two five seven times ten to the power minus three square metres. With me? Put it all together: five point six seven times ten to the power minus eight, times one point two five seven times ten to the power minus three, times the bracket. Clear? About seventy point seven watts, so about seventy one watts. And notice: eight point one times ten to the power nine is tiny next to ten to the power twelve, so the cold walls hardly matter. That's it!

8:04Quick revision: steady state, rate of cooling, net radiation

So, quick revision. One: in the steady state, heater power equals the rate of heat loss. Two: just after switch-off, the loss is the same, so the rate of cooling, minus d T by d t, equals P over m s. Three: the net radiation loss is e sigma capital A, times, T to the power four minus T naught to the power four, in kelvin; never T minus T naught, the whole to the power four. Well done! Now go and try the three practice problems on this page. All the best!