Mixed Problems: Set 2

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. You drop fifty grams of ice, at zero degrees Celsius, into two hundred grams of water at forty degrees. Many students treat the ice as just cold water, take the average of the temperatures, weighted by mass, and get thirty two degrees. But the true answer is exactly half of that! So where did the heat go? We will settle it in example one. Today's mixed set brings elasticity, surface tension, heat and flowing water together, the way JEE questions often do. Let's begin.

0:38Idea 1: a stretched wire, its diameter and volume

So, idea one, from elasticity. Read it with me: a wire with Poisson's ratio zero point three is stretched by zero point two percent. Find the fractional changes in its diameter and in its volume. Okay? Poisson's ratio is sigma, the Greek letter; in this problem, sigma means Poisson's ratio, not stress. Sigma equals minus the lateral strain, sideways, over the longitudinal strain, along the length. Here it is zero point three. Clear? The longitudinal strain, epsilon, another Greek letter, is delta L over L: zero point two percent here. Delta, the little triangle, means change in. Got it? Now, the wire gets longer. Does its diameter grow, or shrink? It shrinks, yes! That is why the minus sign is there. For the diameter, small d, delta d over d is minus sigma times epsilon: minus zero point three times zero point two percent, which is minus zero point zero six percent. Right? Now the volume. Longer by zero point two percent, thinner by zero point zero six. So the volume goes up by zero point one four percent, right? No, no, careful! The volume goes as d squared times L; the curly sign means proportional to, and the double arrow means so. The diameter is squared, so its change counts twice. Delta V over V is epsilon, plus two times minus sigma epsilon, which is epsilon times, one minus two sigma, in brackets. With me? So delta V over V is zero point two percent, times one minus zero point six, that is zero point four: plus zero point zero eight percent. The volume grows a little, because sigma is below one half. That's it!

2:33Idea 2: a soap bubble has two surfaces

Idea two, from surface tension. Read it with me: how much work does it take to blow a soap bubble up from a radius of one centimetre to two centimetres, at constant temperature? Capital T, the surface tension, is zero point zero three newton per metre; the power minus one on the m means per metre. Okay? Surface tension also equals the surface energy per unit area. The work to make new surface is capital T times delta A, the increase in area, for each surface. Right? Now picture the bubble. How many surfaces does a soap bubble have? Two, exactly! The soap film has air on both sides, so there is an outer surface and an inner surface. But a drop of water has only one. This is the classic trap here. Clear? So the work is two capital T, times four pi, times r two squared minus r one squared, in brackets. Four pi r squared, with pi the Greek letter, is the area of a sphere, and r one and r two, with small subscripts, are the two radii. Got it? Now the numbers. Two squared minus one squared is three square centimetres, that is, three times ten to the power minus four square metres. With me? So the work is eight pi, times zero point zero three, times three times ten to the power minus four: about two point two six times ten to the power minus four joules; the wavy equals sign means approximately. Forget the second surface, and you get only half of this. Okay?

4:11Example 1: 50 g ice in 200 g water at 40 C, 16 C not 32 C

Now, remember our puzzle? Here it is, example one. Fifty grams of ice at zero degrees Celsius is added to two hundred grams of water at forty degrees Celsius. Find the final temperature. Pause the video, and try it first. Done? Okay, let's check. First, look at the heating curve: temperature against heat supplied. On the flat parts, melting and boiling, heat goes in but the temperature does not rise, so Q equals m L, okay? On the sloping parts, Q equals m s delta theta, where theta is temperature. Clear? This chapter's standard values: The specific heat of water, s with water written below, is one calorie per gram per degree Celsius, and the latent heat of fusion, L f, f for fusion, is eighty calories per gram. Okay? Now, cool the water from forty degrees down to zero. Two hundred times one times forty: it can give eight thousand calories. Right? To melt all the ice: fifty times eighty, four thousand calories, the flat part of the curve. Got it? Now, before writing any mixing equation, what must we check first? Whether all the ice melts, yes! Eight thousand is greater than four thousand, as the greater-than sign shows, so there is enough heat. If not, the answer would be zero degrees, with ice left over. With me? So four thousand calories are left over, to warm all two hundred and fifty grams of water from zero up to T, the final temperature. Eight thousand minus four thousand equals two hundred and fifty T, so T is sixteen degrees Celsius. Puzzle solved! The melting ate four thousand calories without warming anything. That's it!

6:03Example 2: flow rate through a venturi meter from 1500 Pa

Now a harder one, example two. Here it is: a horizontal venturi meter has areas of ten square centimetres and five square centimetres. The pressure difference between the wide and narrow parts is one thousand five hundred pascals, for water. Find the flow rate. Pause and try. Hint: continuity first, then Bernoulli. Done? Okay, let's check. The figure shows a pipe going from a wide section, one, to a narrow section, two. Capital P is the pressure, capital A the area, v, the red arrow, the speed, and small h the height above the dashed reference line. Okay? Continuity: A one v one equals A two v two. The narrow area is half the wide one, so the water moves twice as fast there: v two equals two v one. Right? Now Bernoulli. But first, its conditions. The fluid must be ideal: incompressible, and non-viscous. Clear? And the flow must be steady and streamline, not turbulent, and we compare two points along the same streamline. Got it? The figure's pipe rises, but our meter is horizontal, so h one equals h two and the rho g h terms cancel. Rho, the Greek letter, is the density. So P one minus P two equals half rho, times v two squared minus v one squared. With me? Put in rho for water, one thousand kilograms per cubic metre, and v two squared as four v one squared. One thousand five hundred equals half times one thousand, times four minus one, times v one squared. So v one is one metre per second. Okay? Now the flow rate, Q, is A one times v one. But first, ten square centimetres in square metres. How much is it? Ten to the power minus three! Remember, one square centimetre is ten to the power minus four square metres, not ten to the power minus two. So Q is ten to the power minus three cubic metres per second, that is, one litre per second, since one litre is ten to the power minus three cubic metres. Perfect!

8:16Quick revision: wire volume, soap bubble, ice melt, venturi

So, quick revision. Stretched wire: the diameter changes by minus sigma epsilon, the volume by epsilon times, one minus two sigma. Soap bubble: two surfaces, so the work is two T delta A, twice that for one surface. Ice in water: first check that all the ice melts. Venturi meter: continuity first, then Bernoulli. Now go and try the practice problems on this page. All the best!