Energy Conversion, Heat and Efficiency

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Picture a car of twelve hundred kilograms, moving at twenty metres per second. The driver brakes, and the car comes to rest. All that kinetic energy, and now the car has none. So where did it go? Energy cannot just disappear, right? The clue is in the brakes: they get hot. But how hot? By the end of this lesson, you will work it out in just three short lines. Energy turning into heat comes up often in JEE problems. So, let's begin.

0:35Energy is never lost

So, back to our car. Before braking, it had kinetic energy, K E for short, and after braking, none at all. Did that energy simply vanish, or did it go somewhere? It went somewhere! Yes, exactly! Here is the first big rule: energy is never lost. It only changes from one form to another. Okay? Now, two things take mechanical energy, that is kinetic plus potential energy, away from a body. First, friction, like the brakes rubbing on the wheels. Second, an inelastic impact: a collision in which kinetic energy is not conserved, like a bullet that stops inside a sandbag. Right? And the energy they remove appears as heat, sound, or deformation, which means a change of shape. The brakes heat up. The bullet heats up, there is a thud, and the sand is pushed out of shape. So in every such problem, first ask: where did the energy go? Clear?

1:37Heat: warming and melting

Now, when that energy becomes heat, a body warms up, and we need to know by how much. To warm a body: Q equals m s delta theta. Capital Q is the heat, m is the mass, and s is the specific heat of the material. And delta theta: theta, the Greek letter, stands for temperature, and the little triangle in front, delta, means change. So delta theta is the rise in temperature. With me? Now think. The specific heat is given per kilogram. So if a question gives the mass in grams, can you put that number straight into the formula? No! Change grams to kilograms first. Keep every quantity in these units: m in k g, kilograms; s in joules per kilogram per kelvin, where the small minus ones on screen mean per; and delta theta in K, kelvin. Okay? Next, to melt a body, the formula is Q equals m times L, where L is the latent heat: the heat needed to melt one kilogram of the material once it is already at its melting point. Its temperature does not rise while it melts. Got it? And one conversion. Cal is the calorie, an older unit of heat, and capital J is the joule. One calorie equals four point two joules. So if heat comes in calories, multiply by four point two to get joules. Right?

3:02Efficiency: how much of it is useful

Now, machines. Your body, a motor, a crane: energy goes in, and only part of it comes out as useful work. The rest is not lost, remember; it becomes other forms, like heat. So we define efficiency, eta: the Greek letter that looks like a small n with a long tail. Eta equals the useful energy out, over the energy in. So sixty percent efficient means eta is zero point six. Clear? Now think. In most problems you know the useful work and the efficiency, and you need the energy in. Should the energy in be more than the useful work, or less? More! Correct! Part of the input always turns into other forms, so you must put in more than you get out. So, energy in equals the useful work, over eta. Dividing by a number less than one makes it bigger, exactly as it should. With me?

4:01Topper's trick: one energy balance

Now, a trick toppers use for every heat problem: three steps, always in the same order. Step one: find the mechanical energy lost. That is K, the kinetic energy, when a moving body stops, or m g h when a body falls through a height h, with g the acceleration due to gravity. Okay? Step two: keep only the fraction that becomes heat. If the question says eighty percent stays as heat, multiply by zero point eight. If it says all of it, keep the whole amount. Right? Step three: set it equal to m s delta theta, and solve for whatever is asked. Energy lost on one side, heat gained on the other. That's it! Got it?

4:44Example 1: energy used by a crane of 60% efficiency

Now, let's use all this. Example one. Here is a crane with sixty percent efficiency. It lifts five hundred kilograms through twelve metres. How much electrical energy does it use? Take g as ten. On screen, k g is kilograms, and the small m is metres. Pause the video, and try it first. Done? Okay, let's check. First, the useful work. The crane's job is to lift the load, so the useful work is m g h. Five hundred times ten times twelve: sixty thousand joules. Clear? Now, the efficiency is sixty percent. So the energy used is zero point six times sixty thousand, right? No, no, careful! That would make the energy in less than the useful work. Impossible! The energy in is always more, remember? So divide: energy in equals the useful work over eta, that is sixty thousand over zero point six. Okay? Sixty thousand over zero point six is one hundred thousand joules. That is one hundred k J, kilojoules. And the energy that did not become useful work is not lost: it turns into forms we cannot use, like heat. Got it?

6:01Example 2: temperature rise of a braking car's brake drums

Now, remember our puzzle from the start? Let's solve it. So: a twelve hundred kilogram car is braked from twenty metres per second to rest. On screen, m s to the minus one means metres per second. All the heat goes into the steel brake drums, of total mass eight kilograms, and their specific heat s is four hundred and eighty joules per kilogram per kelvin. Find their rise in temperature. Pause and try. Hint: use the three steps. Done? Okay, let's check. Step one: the mechanical energy lost. The car stops, so it loses all its kinetic energy. K equals one half times twelve hundred times twenty squared. Twenty squared is four hundred, so that is six hundred times four hundred: two point four times ten to the power five joules. Clear? Step two: all of it becomes heat, so we keep the whole amount. Step three: set it equal to m s delta theta. And the mass m is twelve hundred kilograms, right? No, no, wait! This is the trap. The car's mass gave us the kinetic energy. But the body that heats up is the brake drums: eight kilograms. So two point four times ten to the power five equals eight times four hundred and eighty times delta theta. Okay? Eight times four hundred and eighty is three thousand eight hundred and forty. So delta theta equals two point four times ten to the power five, over three thousand eight hundred and forty: sixty two point five kelvin! See? Puzzle solved. The car's kinetic energy did not vanish; it became heat in the brake drums. Got it?

7:47Quick revision: energy to heat, Q = ms delta T, efficiency

So, quick revision. Energy is never lost: friction and impacts turn mechanical energy into heat, sound and deformation. Heat: Q equals m s delta theta to warm a body, and Q equals m L to melt it; one calorie is four point two joules. Efficiency: energy in equals the useful work over eta, so divide by eta, never multiply. Now go and try the practice problems on this page. All the best!