Specific Heat of a Solid by Mixtures

Transcript

0:00Introduction

Hello everyone, here is a small puzzle. You drop a hot solid into water to find its specific heat, and the experiment tells you to start with the water a few degrees colder than the room. Why would anyone make it colder on purpose? By the end, you will see exactly why. Questions on this method of mixtures do come up in JEE, so let's begin.

0:24The method of mixtures

Look at the figure and follow the arrows: a copper calorimeter with water, a hot solid, a stirrer and a thermometer, inside an insulating jacket with a lid. In an insulated vessel, the heat lost by the hot bodies equals the heat gained by the cold bodies. That is the method of mixtures. Clear? The solid has mass m s, that is m with a small s below it, and it is heated to theta s, with theta, the Greek letter we use here for temperature. It goes into water of mass m w at theta w, a small w for water. Is the water the only thing that gains heat? Very good if you said no, because the vessel warms up as well. So we count the polished copper calorimeter and its stirrer, mass m c, a small c for calorimeter. Got it? To keep the heat inside, it stands on an insulating support, inside a jacket, with a lid, okay? We stir gently and record the highest steady temperature, which is just theta, with nothing written below it. Right?

1:23Writing the heat balance

Now let us write the balance. The solid cools from theta s down to theta. So how much heat does it lose? It is mass times specific heat times the fall: m s times c s, times theta s minus theta, in brackets. Here c s is the specific heat of the solid, the one we want. Okay? The water and the calorimeter both rise from theta w to theta, with specific heats c w and c c. So the heat gained is m w times c w, plus m c times c c, this whole bracket times theta minus theta w. With me? Put the two equal, and this is the working equation, with only c s unknown. Clear?

2:06Water equivalent of the calorimeter

The water equivalent of the calorimeter is the mass of water with the same heat capacity. How would you write this mass in symbols? So the water equivalent, capital W, equals m c times c c, over c w. Then the right side becomes m w plus capital W, in brackets, times c w, times theta minus theta w, okay?

2:30Precautions, and which way the error goes

Now the precautions, with their reasons. The solid is heated in a steam jacket, so that it is dry and at a known temperature. Lifted out of boiling water, it carries hot water with it, and that heat is wrongly credited to the solid. Got it? Suppose the solid is carried slowly across the room. Will the calculated specific heat come out too large, or too small? That is correct if you said too small, because the solid is cooler than the recorded theta s, so the water warms less. Clear? Now suppose some water splashes out. Something was lost, so is the answer too small again? Oh no, be careful, because it is the opposite. Less water is heated than the mass used in the calculation, so the rise is larger than it should be, and the specific heat comes out too large. Right? So the solid must be transferred quickly and without splashing, and now you know why. With me?

3:27Back to our puzzle: why start below room temperature?

Now, back to our puzzle. Once the water is warmer than the room, some heat leaks out to the room. What does this do to the final temperature, and to the specific heat we calculate? Yes, exactly, both come out low, because heat lost to the room makes theta too low, and c s too small. Okay? So start the water a few degrees below room temperature, so that it ends the same amount above. While the water is below the room, does it gain heat or lose heat? It gains heat while it is below the room and loses heat while it is above, so the two roughly cancel. That is why we make the water colder on purpose, and our puzzle is solved.

4:09Example 1: specific heat of a solid in a copper calorimeter

Now the easier example. Here, a solid of mass two hundred grams at one hundred degrees Celsius, the small circle being the degree sign, is dropped into one hundred and fifty grams of water at twenty five degrees Celsius, in a copper calorimeter of mass one hundred grams. The final temperature is thirty three degrees Celsius. The specific heats are four thousand two hundred for water and three hundred and ninety for copper, in joules per kilogram per kelvin, the unit with two small minus ones. Find the specific heat of the solid. Pause the video and try it first. Done? Okay, let's check. The water and the calorimeter rise from twenty five to thirty three, which is eight degrees Celsius. The masses are in grams, so can you multiply straight away? The answer is no, and this is a common slip, because the specific heats are per kilogram, so change to kilograms first. Zero point one five zero times four thousand two hundred, plus zero point one zero zero times three hundred and ninety, all times eight, gives five thousand three hundred and fifty two joules gained. Okay? The solid falls from one hundred to thirty three, which is sixty seven degrees Celsius. So the heat lost is zero point two zero zero, times c, times sixty seven. Right? Heat lost equals heat gained, so c is five thousand three hundred and fifty two, over zero point two zero zero times sixty seven. The wavy equals sign means approximately, and c is about four point zero times ten to the power two joules per kilogram per kelvin, okay?

5:45Example 2: percent error in specific heat from thermometer

Now the harder one. In the experiment, theta s is one hundred point zero degrees Celsius, theta w is twenty two point zero degrees Celsius, and theta is thirty point zero degrees Celsius. Taking the masses as exact, find the percentage error in c s, with a thermometer reading to zero point five degree Celsius in part a, and zero point one degree Celsius in part b. Pause the video and try it first. Done? Okay, let's check. From our equation, c s varies as the rise over the fall, here the rise is eight point zero and the fall is seventy point zero degrees Celsius. Each difference uses two readings, each off by some amount. So how much can the difference be off? Yes, it can be off by twice that, because errors add even when readings subtract. The little triangle is the Greek capital delta, and delta theta means the error in one reading. Fractional errors add too, so delta c over c equals two delta theta over eight point zero, plus two delta theta over seventy point zero. Clear? For part a, delta theta is zero point five, so two delta theta is one. One over eight point zero, plus one over seventy point zero, is about zero point one three nine, so about fourteen per cent. Okay? For part b, delta theta is zero point one, so two delta theta is zero point two. Zero point two over eight point zero, plus zero point two over seventy point zero, is about zero point zero two eight, so about two point eight per cent. Right? Compare the two terms, and the one with eight point zero is far bigger. So the largest error comes from the small rise, and a thermometer reading to zero point one degree Celsius is preferred. Got it?

7:32Quick revision: heat balance, water equivalent, main error

So, quick revision. Heat lost by the solid equals heat gained by the water and the calorimeter. So m s times c s, times the fall, equals m w plus capital W, times c w, times the rise, where capital W is m c times c c, over c w, okay? The water starts below room temperature and ends the same amount above. And the largest error comes from the small rise, so use a thermometer reading to zero point one degree Celsius. You have the whole method now, so go and try the practice problems on this page. All the best!