Specific Heat of a Liquid by Mixtures

Transcript

0:00Introduction

Hello everyone, here is a small puzzle. Drop a hot solid into some water in a calorimeter. Then heat it again, and drop it into an equal mass of oil, at the same starting temperature. Same solid, same mass, same start, so both should warm up equally, right? Well, they do not, and soon you will tell me which one warms more, and why. This method does come up in JEE, so let's begin.

0:28The setup: a liquid in the calorimeter

Look at the arrangement on screen: a copper calorimeter, in red, inside an insulating jacket with a lid, with a thermometer, a stirrer and the hot solid. The label says water, but here the vessel holds our liquid, okay? The line at the bottom is the method: heat lost by the hot body equals heat gained by the cold bodies. In the last section, the specific heat of the solid was the unknown. Now the liquid is the unknown. So let me ask you this: which specific heat must we already know? Yes, exactly, we must already know the specific heat of the solid. So the solid has mass m with a small s below it, specific heat c with a small s, and starts at theta with a small s. Theta, the Greek letter, stands for temperature here, okay? The liquid gets a small l below each letter: its mass, its specific heat, and its starting temperature. That c with a small l is what we are looking for. Clear? The calorimeter gets a small c below its mass and its specific heat. And plain theta, with nothing below it, is the final temperature. With me?

1:37The heat balance

The solid cools from theta s down to theta. So it loses its mass, times its specific heat, times its fall, theta s minus theta, in the round brackets, okay? The liquid warms from theta l up to theta, so it gains its mass, times its specific heat, times its rise, theta minus theta l. Now tell me, is the liquid the only thing that warms up in there? No, it is not, and forgetting this is a common mistake. The calorimeter warms through the same rise, so it gains its own mass, times its own specific heat, times that rise. Right? Now heat lost equals heat gained, with the liquid's term and the calorimeter's term added on the right, times the common rise. Write this down, and we solve it for c with a small l. Okay?

2:25When can we use it?

Now, this equation comes with conditions. The solid must not react with the liquid, or dissolve in it, or our balance no longer contains only these terms. Got it? There is also a second way: we pour the warm liquid into cold water in the calorimeter. But this is possible only if the two do not react, and no heat is released or absorbed on mixing. Clear? Now tell me, can we find the specific heat of ethanol by mixing warm ethanol with cold water? No, we cannot, because heat is released when these two liquids mix. So the heat balance no longer contains only the terms of the method of mixtures.

3:05Low specific heat: extra care

Right? Now let us settle our puzzle. The same hot solid goes into equal masses of water and oil, at the same temperature. So which one is warmed more, and why? Yes, exactly, it is the oil, because its specific heat is lower, so a given amount of heat produces a greater rise. If you said the oil, very good, our puzzle is settled. Any liquid of low specific heat shows a large temperature change, but it also loses heat to the room faster, okay? So the precautions of the last section matter even more here: the lid and the jacket, a quick transfer, and gentle stirring. Got it? Now here is a trap. When the solid was the unknown, heat lost to the room made the answer too small. So now, with a known hot solid dropped into the liquid, does a heat leak make the calculated value too large or too small? It comes out too large, the opposite of the last section, so do not just copy the earlier result. The leak makes the final temperature lower than the ideal one, so the liquid seems to have needed more heat per degree.

4:14Example 1: specific heat of oil with a hot copper block

Clear? Let us try the easier example first. Here, a copper block of one hundred grams at one hundred degrees Celsius, the small circle meaning degrees, goes into one hundred and twenty grams of oil at twenty five degrees. The copper calorimeter has mass sixty grams, and the final temperature is thirty seven degrees. Find the specific heat of the oil. Copper has three hundred and ninety joules per kilogram per kelvin, with capital J for joule, capital K for kelvin, and the minus one powers simply meaning per. Pause the video and try it first. Done? Okay, let's check. The block falls from one hundred to thirty seven, which is sixty three degrees. The oil and the calorimeter rise from twenty five to thirty seven, which is twelve degrees. With me? Specific heat is per kilogram, so use kilograms. The block loses zero point one, times three hundred and ninety, times sixty three, which gives two thousand four hundred and fifty seven joules, okay? Careful now, not all of this reaches the oil. The copper calorimeter takes zero point zero six, times three hundred and ninety, times twelve, which gives about two hundred and eighty one joules. So the oil gets the difference: two thousand four hundred and fifty seven minus two hundred and eighty one, which gives two thousand one hundred and seventy six joules. Its own term is zero point one two, times c, times twelve, with c its specific heat. Got it? So c is that number, over zero point one two times twelve. The wavy equals sign means approximately, so c is about one point five times ten to the power three joules per kilogram per kelvin. Clear?

5:55Example 2: liquid's specific heat without the solid's mass

Now the harder one. Here, a solid at one hundred degrees raises one hundred grams of water from twenty to twenty eight degrees. Heated again to one hundred, it raises one hundred grams of a liquid from twenty to thirty five. Neglecting the calorimeter, find the specific heat of the liquid. Pause the video and try it first. Done? Okay, let's check. The question gives neither the mass of the solid nor its specific heat. So how can we use the heat balance at all? We do not need them separately, which is the neat trick. Mass times specific heat is one number, the heat capacity, and it is the same solid in both drops. Okay? First the drop into water, with specific heat four point two joules per gram per kelvin. The water rises by eight degrees, and the solid falls from one hundred to twenty eight, which is seventy two. So the heat capacity of the solid is one hundred times four point two times eight, over seventy two, which gives forty six point seven joules per kelvin. Clear? Now the drop into the liquid. It is tempting to use seventy two again. So tell me, how far does the solid fall this time? It falls only sixty five degrees, because now it stops at thirty five, and that was the trap. The liquid rises from twenty to thirty five, which is fifteen degrees. Got it? With no calorimeter term, forty six point seven times sixty five equals one hundred times c times fifteen, and c comes to two point zero joules per gram per kelvin. That is lower than four point two for water, which is why the liquid rose by fifteen and the water by only eight. That is our opening puzzle again, right?

7:44Quick revision: heat balance, conditions and heat leak error

So, quick revision. Heat lost by the solid equals heat gained by the liquid and the calorimeter. No reaction or dissolving, and no heat released or absorbed on mixing. Low specific heat means a bigger rise and a faster loss, and with a hot solid, a heat leak makes the answer too large. Now go and try the practice problems on this page. All the best!