Past JEE Questions: Heat and Sound Experiments

Transcript

0:00Introduction

Hello everyone, here is a small puzzle from an old paper. In winter, a student finds the first resonance of an air column at eighteen centimetres. In summer she repeats the experiment and measures the second resonance. Now you might say the second resonance is three times the first, so fifty four centimetres, right? But two separate things are wrong with that answer, and we will catch both today. This part of the course has three past JEE questions on heat and sound, so these experiments do come up. So, let's begin.

0:33Heat: the balance in a calorimeter

We will come back to that puzzle, but let us start with heat, so look at the figure. The hot solid sits in a copper calorimeter with water, a thermometer and a stirrer, inside an insulating jacket with a lid. This is the method of mixtures. In an insulated vessel, the heat lost by the hot body equals the heat gained by the cold bodies, okay? Small m is a mass, small c is a specific heat, and theta, this Greek letter, is a temperature. The small s below each letter means the solid, so theta with that s is where it starts, and plain theta is the final steady temperature. So the heat lost is mass, times specific heat, times the fall, which is the bracket. Right? So who gains this heat, only the water? No, the calorimeter gains it too, and forgetting it is the classic mistake. Here small w means water, and a small c below a letter means calorimeter. Each takes its mass times its specific heat, we add the two, and we multiply by the rise, the last bracket, theta minus theta w, with me?

1:39Sound: the first two resonances

Now for sound, look at this new figure. The red curve is the stationary wave in the air column. On the left is the first resonance, and the arrow marks l one, the length from the open end down to the water, with a small one below the l as a label. The dashed lines at the top mark small e, the end correction, because the antinode lies a little above the open end. With lambda, this Greek letter, as the wavelength, l one plus e equals lambda over four, okay? On the right is the second resonance, with length l two, and there l two plus e equals three times lambda, over four. This figure is only a sketch, so its lengths are not to scale. Clear? Now let us check our puzzle. Is l two exactly three times l one? No, it is not, because small e sits in both lines. It is l two plus e that equals three times the bracket, l one plus e, so l two is not three times l one. That was the first mistake in our puzzle. Got it? The figure notes add two more results. Small v, the speed of sound, equals two times small f, the fork frequency, times the bracket, l two minus l one. And small e equals l two minus three times l one, all over two, and the wavy sign says that is roughly zero point three times small d, the internal diameter.

3:01Sound: what a warmer room changes

Okay? Now let us bring in the temperature. Small v is proportional to the square root of capital T, the absolute temperature, and the curly symbol means proportional to. Clear? Now keep the same fork, so small f stays fixed, and lambda equals small v over small f. So on a warmer day, what happens to the wavelength? It becomes greater, and that is the second mistake in our puzzle. Greater speed with small f fixed means a greater wavelength, right?

3:33Example 1: specific heat of a metal in a brass calorimeter

Now Example one, from JEE Main twenty nineteen. An unknown metal of mass one hundred and ninety two grams, heated to one hundred degrees Celsius, is immersed in a brass calorimeter of mass one hundred and twenty eight grams, containing two hundred and forty grams of water at eight point four degrees Celsius. The water temperature settles at twenty one point five degrees Celsius. Find the specific heat of the metal. Brass is three hundred and ninety four, and water is four point one eight times ten to the power three, both in joules per kilogram per kelvin, as the powers of minus one on screen show. Pause the video and try it first. Are you done? Okay, let us check it together. The water and the brass rise by twenty one point five minus eight point four, which gives thirteen point one degrees Celsius. Got it? The masses are in grams, but the specific heats are per kilogram. So what must we do first? Yes, exactly, we change the masses to kilograms. Now four point one eight times ten to the power three is just four thousand one hundred and eighty. So the water gives zero point two four zero times that, and the brass gives zero point one two eight times three hundred and ninety four. Add the two, multiply by thirteen point one, and we get about thirteen thousand eight hundred joules, okay? Now take the metal, which started at one hundred degrees. It falls to which temperature? It falls to twenty one point five degrees, the final temperature, and taking eight point four here is an easy slip. So the fall is one hundred minus twenty one point five, which gives seventy eight point five degrees Celsius. Right? The heat gained equals the heat lost. So thirteen thousand eight hundred equals zero point one nine two, times small c, times seventy eight point five, where small c is the specific heat of the metal. Clear? Now divide, and small c comes to about nine hundred and sixteen joules per kilogram per kelvin. That is our answer, and very good if you remembered the brass.

5:39Example 2: second resonance in summer, 18 cm in winter

Now Example two, the harder one, and it is our puzzle, from A I triple E two thousand and eight. While measuring the speed of sound with a resonance column, a student gets the first resonance at a column length of eighteen centimetres during winter. Repeating the experiment in summer, she measures the column length for the second resonance, and x, in centimetres, is that length. What can we say about x, this summer length? Pause the video and try it first. Are you done? Okay, let us check it together. The first resonance in winter is at eighteen centimetres, so eighteen plus small e equals lambda over four, with small e still the end correction, okay? Now the second resonance is measured in summer, with the same fork and the same tube. So is the summer wavelength the same as the winter wavelength? No, and that is exactly the trap. In summer the speed, and so the wavelength, is greater. Call it lambda prime, lambda with the small dash on screen, and the pointed sign means greater than. Right? So small x plus e equals three times lambda prime, over four. That is greater than three times lambda, over four, and from the winter line, that equals three times the bracket, eighteen plus e, clear? Open the bracket, and the right side becomes fifty four plus three times e. Take e away from both sides, and small x is greater than fifty four, plus two times e, got it? And small e is a positive length, so small x is greater than fifty four centimetres. So fifty four was wrong twice over, because of the end correction and because of the summer air. If you caught both on your own, very good, that is really careful physics.

7:28Quick revision: heat balance, end correction, warmer air

So, let us do a quick revision. The heat lost by the hot body equals the heat gained by the water and the calorimeter. Then, l one plus e equals lambda over four, and l two plus e equals three times lambda, over four, so l two is not three times l one. And warmer air with the same fork means greater speed and a greater wavelength. One more past question waits on this page, a metal ball in a vessel of water. It uses this same heat balance, so try it before reading its solution, and all the very best to you.