Overview

Transcript

0:00Introduction

Hello everyone! Here is a puzzle from the heat lab. We drop a hot solid at ninety five degrees Celsius into water at twenty degrees, and it all settles at just twenty five. The solid cooled by seventy degrees, but the water warmed by only five. Did most of the heat vanish? Hold that thought; we settle it in Example two. Let's begin.

0:24How this chapter works

So, here is how this chapter works: three timed sets of fifteen original problems each, in the style of recent JEE Main papers. Okay? Each set mixes all the experiments of the unit, so first spot the experiment. Right? Give yourself twenty five minutes per set, then check the solutions. So how many seconds does each problem get? Twenty five minutes is one thousand five hundred seconds, the small s, and the division sign means divide by fifteen, giving one hundred seconds per problem. So keep the tools ready in your head. Clear? Unless told otherwise, take g, the acceleration due to gravity, as nine point eight metres per second squared, m s to the power minus two. With me? For water, take the density as one thousand kilograms per cubic metre, k g m to the power minus three, and the specific heat as four point two joules per gram per kelvin, J g to the power minus one, K to the power minus one. Got it?

1:28Tools 1: vernier callipers and screw gauge

Now, the tools for the vernier callipers and the screw gauge. L C, the least count, is one main scale division, MSD, minus one vernier scale division, VSD, okay? If capital N vernier divisions equal N minus one main scale divisions, in brackets, the least count is one MSD over capital N, right? The vernier reading is the main scale reading, MSR, plus small n, the coinciding vernier division, times the least count; the cross means times. Then subtract the zero error. Okay? For the screw gauge, the least count is the pitch, the advance in one rotation, over capital N, the number of circular scale divisions. The reading is MSR, plus CSR, the circular scale reading, times the least count, minus the zero error. Clear? Now, a classic trap. The zero error is always subtracted with its sign. So if it is negative, do you add it or subtract it? You add it, because minus a negative number is a plus. So a positive zero error is subtracted, and a negative one is added. Got it?

2:38Tools 2: errors and the mechanics experiments

Next, errors and the mechanics experiments. For a product or a quotient, the fractional errors add. Delta, the little triangle, means the error, so delta x over x is the fractional error, and x to the power n brings n times it. Okay? For the simple pendulum, g equals four pi squared times L, over T squared, with pi the Greek letter and T the time period. Careful here: the length L runs from the point of suspension to the centre of the bob, so it is the thread, plus the hook, plus the radius of the bob. Right? For Young's modulus, capital Y equals four times M times g times capital L, over pi times d squared times small l, where M is the load, capital L the length, d the diameter and small l the extension. Which error counts twice in Y? Yes, exactly, the diameter, because it enters squared. So delta Y over Y equals delta L over L, plus two times delta d over d, plus delta l over l. The diameter's error usually dominates.

3:47Tools 3: sound, heat and electricity

Clear? Now, the third set of tools. In the resonance tube, v, the speed of sound, equals two f times l two minus l one, in brackets, with f the frequency of the fork, and l one and l two, small one and two below, the first two resonance lengths. No end correction is needed. Okay? The end correction itself, e, is l two minus three times l one, all over two. With me? For mixtures, heat lost equals heat gained, and the calorimeter gains heat too; forgetting it is the common trap. Right? Here small m is a mass, small c a specific heat, and theta, the Greek letter like an O with a bar, a temperature; subscript s means solid, w water, theta alone is the final temperature, and capital W is the calorimeter's water equivalent. So the solid's m c times its fall equals m w plus capital W, times c w, times the water's rise. Clear? In the metre bridge, capital R is the resistance in the left gap and capital S the one in the right gap, and R over S equals l over one hundred minus l, with l the balancing length in centimetres from the left end. First check which gap holds the unknown; swapping them moves the null point from l to one hundred minus l. Got it?

5:08Example 1: vernier reading when 20 VSD equal 19 MSD

Alright, Example one, from Set one. Twenty vernier divisions equal nineteen main scale divisions of one millimetre, m m, each. The main scale reads four point one centimetres, c m, and the thirteenth vernier division coincides. Find the reading. Pause the video and try it first. Done? Okay, let's check. The least count is one millimetre over twenty, which gives zero point zero five millimetres. Now, can we add thirteen times zero point zero five straight to four point one? No, not directly, because the main scale reads in centimetres. So convert first: zero point zero five millimetres is zero point zero zero five centimetres. Mixing units is a common slip in a timed set. Right? So the reading is four point one, plus thirteen times zero point zero zero five, that is, plus zero point zero six five, giving four point one six five centimetres, well inside our hundred seconds. Clear?

6:15Example 2: specific heat of a solid by method of mixtures

Now Example two, also from Set one, and it is our puzzle. We drop a solid of mass eighty grams, the small g, at ninety five degrees Celsius, the small circle with capital C, into one hundred grams of water at twenty degrees Celsius, in a calorimeter of water equivalent twenty grams. The final temperature is twenty five degrees Celsius. Find the specific heat of the solid. Pause the video and try it first. Done? Okay, let's check. The figure shows the hot solid in a copper calorimeter of water, with a thermometer and a stirrer, inside an insulating jacket with a lid. Heat lost equals heat gained, as the figure says. So, besides the water, who gains heat here? Yes, exactly, the calorimeter, so its twenty grams of water equivalent join the one hundred grams of water in the first brackets. And the rise, twenty five minus twenty, sits in the second brackets. Okay? So, one hundred and twenty times four point two, times five, gives two thousand five hundred and twenty joules, the capital J, gained. Right? The solid loses heat: eighty grams, times its unknown specific heat c, times its fall, ninety five minus twenty five, which is seventy. Eighty times seventy gives five thousand six hundred, so it loses five thousand six hundred c. With me? Heat lost equals heat gained, so five thousand six hundred c equals two thousand five hundred and twenty; the double arrow means so. Then c is two thousand five hundred and twenty over five thousand six hundred, which gives zero point four five joules per gram per kelvin. Clear? So our puzzle is solved, and no heat vanished. The water and the calorimeter gained exactly the two thousand five hundred and twenty joules the solid lost; they just need far more heat per degree. Got it?

8:16Quick revision: zero error, error rules, heat balance

So, quick revision. Subtract the zero error with its sign, and match units before adding. Fractional errors add, and a squared quantity counts twice. In mixtures, heat lost equals heat gained, calorimeter included. Now open Set one, give it twenty five minutes, then check the solutions. All the best!