Topic 6.4: Heat Capacity and Calorimetry

Heat capacity and calorimetry

It’s Christmas weekend in 2022, it’s the middle of the night in an orange grove in central Florida, and the temperature is sliding below freezing. A night like this can kill a young orange tree, so the growers walk out into the dark and do something that sounds completely backwards. They turn on the water and aim it right at the trunks of their young trees, and they leave it running all night long, until every one of those trunks is wrapped in ice.

And it works. As the night goes on, the air falls into the twenties, but the bark under that ice stays right around thirty-two degrees, the freezing point, which is warmer than the air all around it.

So how can wrapping a tree in ice keep it warm, when ice is exactly what the cold is trying to make? By the end of this lesson, you’ll be able to work out exactly how much heat that water hands the tree, on its way to becoming ice.

And you’ll also know why, on one windy night in 1962, trees that growers sprayed with water died anyway, killed by the same water that’s saving these ones. That water comes up out of a well at about sixty-eight degrees Fahrenheit, which is twenty degrees Celsius, so before it can freeze, it has to cool all the way down to zero on the bark, and that’s where we’ll start.

Before we start, make a guess. On the bark, that water first cools all the way from twenty degrees down to zero and then freezes, so which of those two steps do you think hands the tree more heat?

Heat and q = mcΔT

Let’s draw a circle around just the water on that trunk and call it our system, and everything else, the bark, the ice underneath and the night air, we’ll call the surroundings. The water is warmer than the bark, so energy flows out of the water and into the bark, and energy that moves because one thing is hotter than another is called heat, written with the letter q.

When a system cools down, its energy goes down, so we give its q a minus sign, and when something warms up, its energy goes up, and its q is positive. No energy appears or disappears along the way, which is the first law of thermodynamics, so every joule the water loses, the surroundings gain.

How much a temperature changes depends on what you’re heating, and an experiment shows it best. Here are two cups, one with twenty-five grams of water and one with a twenty-five gram block of copper, each with the same little heater. We’re going to give each of them exactly the same energy, five hundred joules, starting from the same temperature.

Before we switch them on, make a prediction. Which one do you think ends up with the bigger temperature rise? Same energy, same mass, and the copper’s temperature changed about eleven times as much as the water’s. Water takes a lot of energy to change its temperature, and copper takes very little, and that exact comparison shows up on the AP exam.

So heating something depends on how much of it there is, what it’s made of, and how far its temperature moves. Put them together and you get q equals m c delta T, where m is the mass in grams, c is the specific heat, the energy it takes to warm one gram of a substance by one degree, and delta T is how much the temperature changes.

Here are a few specific heats, and water’s is huge, four point one eight four joules per gram per degree, about eleven times copper’s, which is exactly what our two cups just showed us. You’ll use this equation on the AP exam again and again, and in 2024 a whole free-response question was one hot metal cube, a cup of water and a thermometer, which we’ll work in a few minutes.

The one thing to watch is delta T, which is always the final temperature minus the starting one, so when something cools, delta T is negative, and q comes out negative all by itself. Let’s do our grove water, following one hundred grams of it, a little under half a cup, as it cools from twenty degrees down to zero.

In go the mass, the specific heat and the change in temperature, zero minus twenty, and the grams cancel, the degrees cancel, and we’re left with joules, which is exactly what heat should come out in. That’s minus eight thousand three hundred sixty-eight joules, or minus eight point three seven kilojoules, and the minus sign says the water lost that energy, straight into the bark and the air around it.

Now you try one with a different substance. A fifty-gram piece of copper pipe out in that grove cools from twenty degrees down to minus three. What’s q for the copper, in joules?

Grams or moles: molar heat capacity

Chemists love counting in moles, so there’s a version of this equation that counts the water in moles instead of grams. Our hundred grams of water is a hundred divided by eighteen point zero two grams per mole, which is five point five four nine moles.

Instead of the energy per gram we use the energy per mole, called the molar heat capacity, about seventy-five point four joules per mole per degree for water, which is just the specific heat times the molar mass. Now q equals n C m delta T, and this time it’s the moles that cancel, and we land on minus eight point three seven kilojoules again, because we’ve counted the same water a different way.

People get stuck trying to remember where the molar mass goes, but the units tell you, because joules per gram goes with grams and joules per mole goes with moles. If a heat capacity comes in joules per kelvin, that’s fine, because a change of one degree Celsius is the same size as a change of one kelvin, so you never add two seventy-three to a temperature change.

Here’s one in moles. One and a half moles of liquid water warms up by twelve degrees, and its molar heat capacity is seventy-five point four. How much heat does it absorb, in joules?

Calorimetry: heat lost = heat gained

Somebody had to measure every one of those specific heats, and the instrument that does it is surprisingly simple, a calorimeter, which is really just an insulated cup of water with a lid and a thermometer. Here’s a fifty gram copper cylinder that’s been sitting in boiling water, so it’s at a hundred degrees, and we’ll drop it into fifty grams of water at seventeen degrees, the same mass as the copper.

The copper’s temperature plunges and the water’s creeps up, until they meet at one temperature and stay there, twenty-four point zero degrees. If you guessed they’d meet halfway, around fifty-eight degrees, that’s a natural guess, and it’s way off. The copper dropped seventy-six degrees while the water rose only seven, about eleven times less, which is our two cups from before, seen from the other side.

The cup is insulated, and if we ignore the little bit of heat the cup itself soaks up, every joule the copper lost went into the water, so the copper’s q and the water’s q add up to zero. I keep the copper’s numbers in one column and the water’s in another, because the classic mistake is grabbing the wrong mass or the wrong starting temperature, and the only number that truly belongs to both of them is the final temperature.

For the water, fifty grams times four point one eight four times seven degrees is about plus fifteen hundred joules, and for the copper, fifty grams times zero point three eight five times minus seventy-six is about minus fifteen hundred joules, the same energy with the opposite sign.

Now let’s do the real one. On the 2024 AP exam, a student heats a ninety-eight point one gram cube of an unknown metal in boiling water, to a hundred degrees, drops it into a calorimeter, and the water tops out at thirty-eight point five degrees. Then it tells you the metal handed two thousand nine hundred forty joules to the water, and it asks for the metal’s specific heat.

Before any math, let’s make sure we grab the right numbers. Tap the three numbers that belong to the metal, its mass and its two temperatures. Now you finish it. What’s the specific heat of the metal, in joules per gram per degree?

The question’s next part asks us to imagine ninety-eight point one grams of aluminum giving off that same energy instead. Aluminum’s specific heat is bigger, zero point eight nine seven, so its temperature moves only about thirty-three degrees instead of sixty-one, because a bigger specific heat means a smaller change. And if you leave the lid off in a real lab, some heat escapes into the room, so the water warms less, and you’d underestimate the heat the metal gave off.

Dissolving: hot packs and cold packs

A calorimeter can even show you a salt dissolving. This is calcium chloride, the salt in last lesson’s hot pack, and as it dissolves, the hundred grams of solution warms from twenty-two degrees to twenty-six and a half. And this is ammonium nitrate, the salt in an instant cold pack, and this time the solution cools, from twenty-two degrees down to eighteen point six.

The trick is knowing whose q you mean. The system is the dissolving itself, and the water around it is the surroundings, which is also our thermometer. With calcium chloride the solution gained energy, a hundred grams of solution times its specific heat, four point one eight, times four and a half degrees, plus eighteen hundred eighty-one joules, so the dissolving gave that energy away, its q is minus one point eight eight kilojoules, and it’s exothermic.

For the ammonium nitrate it runs the other way, because the solution lost fourteen hundred twenty-one joules, so the dissolving took that energy in, plus one point four two kilojoules, and it’s endothermic. The trap is to see the water get hotter and write a positive q for the dissolving, when the water getting hotter is exactly the evidence that the dissolving gave its energy away.

Your turn to read a thermometer like a chemist. A different salt dissolves in a calorimeter, and the solution’s temperature drops. Which of these describes the dissolving?

Three ways a system’s energy changes

Step back for a moment, because a system’s energy changes in three main ways, by heating or cooling, like our grove water dropping from twenty to zero, by changing phase, like that water freezing, or by a chemical reaction, like the iron in last lesson’s hand warmer reacting with oxygen, and reactions get their own lessons next. Our grove water has done the first of those, giving the bark and the air eight point three seven kilojoules for every hundred grams, and at zero degrees it starts the second, which is where the thermometer does something that looks impossible.