Overview

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. In a heat experiment, the final temperature is twenty nine point five degrees Celsius, but someone reads thirty point zero. Half a degree in about thirty is under two per cent, so the answer should be off by about that much, right? But the specific heat comes out about ten per cent higher. How can half a degree do that? Today we see how to use this problem bank, and settle this puzzle with one of its problems. This does come up in JEE, so let's begin.

0:32How to use this chapter

So, here is what this chapter holds. Part A has twenty solved problems at JEE Main level. Okay? Then Part B has nineteen harder, multi-step problems. With me? And Part C has seventeen concept checks on the ideas behind the experiments, then fourteen practice problems. Right? Now tell me, should you read a solution straight away, or only after trying the problem yourself? Yes, exactly, only after trying it. Attempt each part as a test, then read the solutions. Clear?

1:06Values to take, unless told otherwise

Next, the values to take unless told otherwise. First, g, the acceleration due to gravity, is nine point eight metres per second squared, the small minus two meaning per second squared. Okay? The density of water is one thousand kilograms per cubic metre, the small minus three meaning per cubic metre. Got it? And the specific heat of water is four point two joules per gram per kelvin, where each small minus one means per. If your masses are in kilograms, can you still use four point two? You cannot, because it is per gram. Per kilogram, the same value is four thousand two hundred joules per kilogram per kelvin. So match the unit of mass, or you are off by a factor of a thousand. Right?

1:53Part C: why add the water equivalent?

Now one concept check from Part C, on the method of mixtures. In the figure, a thermometer and a stirrer dip into water in a copper calorimeter, with a hot solid at the bottom, all inside an insulating jacket with a lid. In an insulated vessel, heat lost by the hot body equals heat gained by the cold bodies. Okay? Part C asks why the water equivalent of the calorimeter is included in the heat balance. Is the water the only thing that gets warm? It is not, because the calorimeter and the stirrer are warmed through the same range as the water. With me? So they take a share of the heat, which we must count. Clear? For this we use the water equivalent, capital W, the mass of water with the same heat capacity. Capital W equals m c times c c, over c w, where m c, m with a small c below, is the calorimeter's mass, c c its specific heat, and c w the specific heat of water. Got it? Then the heat gained is m w plus capital W, times c w, times theta minus theta w. Here theta, the Greek letter, is the final temperature, theta w the starting temperature of the water, and m w the mass of water. Right?

3:07Example 1: final temperature of hot brass in water

Now Example one, an easier one from Part A, to warm up. Here, a two hundred and fifty gram brass block, of specific heat zero point three eight joules per gram per kelvin, is at one hundred degrees Celsius; the small circle is the degree sign. It goes into two hundred grams of water at twenty two degrees Celsius, in a calorimeter of negligible heat capacity. Find the final temperature. Pause the video and try it first. Done? Okay, let's check. The block is at one hundred and the water at twenty two, so the final temperature is simply the middle, sixty one degrees Celsius, right? That is a trap, because the middle is right only when both sides have the same heat capacity. The brass loses two hundred and fifty, times zero point three eight, times one hundred minus theta, which is ninety five times one hundred minus theta, where theta is the final temperature. Okay? The water gains two hundred, times four point two, times theta minus twenty two, which is eight hundred and forty times theta minus twenty two. The calorimeter's heat capacity is negligible, so it takes no share. Right? Heat lost equals heat gained, so we put the two equal. Clear? Open the brackets, and nine thousand five hundred minus ninety five theta equals eight hundred and forty theta, minus eighteen thousand four hundred and eighty. So nine hundred and thirty five theta equals twenty seven thousand nine hundred and eighty. With me? So theta is twenty seven thousand nine hundred and eighty over nine hundred and thirty five, which gives about twenty nine point nine degrees Celsius; the wavy equals sign means approximately. That is far below sixty one, since the water's eight hundred and forty is much bigger than ninety five. Got it?

4:51Example 2: specific heat, and a 0.5 degree misread error

Now Example two, a harder one from Part B, and it is our puzzle. Here, a one hundred and fifty gram solid at one hundred degrees Celsius is dropped into two hundred grams of water at twenty four point zero degrees Celsius, in a calorimeter of water equivalent fifteen grams. The final temperature is twenty nine point five degrees Celsius. Find the specific heat, and the percentage change in it if the final temperature is misread as thirty point zero degrees Celsius. Pause the video and try it first. Done? Okay, let's check. The water is two hundred grams, and the calorimeter's water equivalent is fifteen grams. So what mass goes into the heat gained? Two hundred plus fifteen, that is two hundred and fifteen grams, just as our concept check said. The rise is twenty nine point five minus twenty four point zero, which is five point five, so the heat gained is two hundred and fifteen, times four point two, times five point five. Okay? The solid falls from one hundred to twenty nine point five, which is seventy point five, so the heat lost is one hundred and fifty, times c, times seventy point five, where c is the specific heat. Right? Equate them, and c is two hundred and fifteen times four point two times five point five, over one hundred and fifty times seventy point five. That gives zero point four seven zero, about zero point four seven joules per gram per kelvin. Got it? Now the misreading, thirty point zero instead of twenty nine point five, so half a degree off. Before you calculate, make a guess. How much will c change? The rise becomes six point zero, and the fall becomes seventy point zero. In the same fraction, c now comes out as zero point five one six. With me? And zero point five one six over zero point four seven zero is about one point one zero. So the result is about ten per cent higher, and our puzzle is settled. Clear? But why did half a degree matter so much? It changed the rise from five point five to six point zero, about nine per cent, but the fall from seventy point five to seventy point zero, under one per cent. So the small rise carries almost all the error. That is why the largest error in this experiment comes from the small temperature rise. Got it?

7:13Quick revision: heat balance, water equivalent, errors

So, quick revision. Part A is JEE Main level, Part B is harder and multi-step, and Part C checks the ideas; test yourself first. Unless told otherwise, g is nine point eight, the density of water one thousand, and its specific heat four point two joules per gram per kelvin. Heat lost equals heat gained, with capital W added to the mass of water. The small rise gives the largest error, so a thermometer reading to zero point one degree Celsius is preferred. Now try the practice problems on this page. All the best!