Constant-pressure calorimetry | Thermodynamics | AP Chemistry | Khan Academy

Calorimetry refers to the measurement of heat flow, and a device that's used to measure heat flow is called a calorimeter. An easy way to make a calorimeter is to use two coffee cups. So at the base here, we have one coffee cup, and then we can also use another coffee cup as a loose fitting lid. Since this top coffee cup is loose fitting, our calorimeter is exposed to the constant pressure of the atmosphere. Therefore, we could use this coffee cup calorimeter for constant pressure calorimetry.

Other components of our calorimeter include some water, and then we also have a stir bar to stir up the water and a thermometer to measure the temperature change of the water. Let's say we have 150.0 grams of water at an initial temperature of 25.0 degrees Celsius. Next, let's take a block of copper, 120.0 grams of it, and let's heat up that block of copper to 100.0 degrees Celsius. Once the copper has reached that temperature, we add the copper block to our calorimeter.

Here we can see the copper block has been added to the calorimeter, and since the copper is at a higher temperature than the water, heat flows from the copper block to the water. Therefore, the temperature of the water will increase, which we will see on the thermometer. We'll see the temperature increase on the thermometer. Heat is transferred from the copper block to the water until thermal equilibrium has been reached. We know when thermal equilibrium has been reached by looking at the thermometer and measuring the highest temperature that's reached.

Let's say the final temperature is equal to 30.0 degrees Celsius. So at thermal equilibrium, both the piece of copper, both the copper block and the water are at the same final temperature. Next, let's calculate the heat gained by the water by using the equation q = mcΔt. So q is what we're trying to calculate, the heat gained by the water.

m is the mass of the water, which is 150.0 grams, so we can write in 150.0 grams. C is the specific heat of water, which is 4.18 joules per gram degree Celsius, and Δt is the change in the temperature, which would be the final temperature (tf) minus the initial temperature (ti). The final temperature of the water is 30.0 degrees Celsius, and the initial temperature of the water was 25.0 degrees Celsius. So 30.0 minus 25.0 is equal to 5.0 degrees Celsius.

So we can write that in, and next we look at units and see what cancels out here. The grams cancel out, degrees Celsius cancels out, and we're left with joules as our unit. So q is equal to, when we go to two significant figures, this is positive 3.1 times 10 to the third joules. The positive sign means that this was the energy gained by the water.

Next, let's do the same calculation for copper. So we're trying to find q. The mass of the copper was 120.0 grams, so we can plug that in. The specific heat of copper is 0.39 joules per gram degree Celsius. Let's think about the change in the temperature of the copper. The final temperature of the copper was 30.0 degrees Celsius, and the initial temperature of the copper was 100.0 degrees Celsius.

So the change in the temperature would be 30.0 minus 100.0, which, of course, is negative 70.0. So let's plug in negative 70.0 degrees Celsius. Once again, we see what cancels for our units. Grams will cancel, degrees Celsius will cancel, and our answer will be in joules.

So q is equal to, using two significant figures, negative 3.3 times 10 to the third joules. The negative sign means this is the energy that was lost by the copper. Next, let's look at these two numbers that we got from our calculations. Let's think about the magnitude of these two numbers. If our coffee cup calorimeter were a perfect insulator, the magnitude of these two numbers would be the same.

So it could be something like 3.3 times 10 to the third joules for both of them. But since these two numbers are not the same, we can see that we've lost more heat from the copper than we've gained in terms of energy for the water, which means we could have lost some of the energy to the environment. So not all of the heat was transferred directly to the water; some of it could have escaped our coffee cup calorimeter.

Next, let's think about calorimetry for a chemical reaction. Before we do that, let's review some terms for thermodynamics. The system is the part of the universe that we are studying. So in the case of a chemical reaction, the reactants and the products make up the system. The surroundings are everything else, which would include the water in the calorimeter, the coffee cup itself, the thermometer, and the environment outside. So the surroundings are everything else.

Finally, the universe would be the system plus the surroundings. The reactants and the products make up the system; so that's what the "s" stands for here in our calorimeter— that's our system. Let's say we run a reaction, and in the reaction, heat is given off. In that case, heat would flow from the system to the surroundings, and so the temperature of the water would increase. We would see that as the temperature increases on the thermometer.

Next, we could calculate the heat gained by the water by using our q = mcΔt equation. Let's say q is equal to positive 1.0 times 10 to the second joules. The positive sign means that the water gained energy. If we assume a perfect transfer of heat from the system to the surroundings, if the surroundings gained positive 1.0 times 10 to the second joules, that means the system must have lost negative 1.0 times 10 to the second joules.

So the same magnitude, but we changed the sign here because if we're talking about the energy lost by the system, it's the same in magnitude but opposite in sign. Next, remember that our lid over here is loose fitting, which makes this constant pressure calorimetry. Therefore, this heat that was transferred is the heat that’s transferred at constant pressure.

So we can write a subscript p in here. So qp, the heat transfer to constant pressure, is the definition for the change in the enthalpy, Δh. So we can write that qp = Δh. When Δh is negative, we're talking about an exothermic reaction. When a reaction is exothermic, heat is transferred from the system to the surroundings, and therefore we see an increase in the temperature of the water.

Finally, let's think about an endothermic reaction. In an endothermic reaction, heat is transferred from the surroundings to the system. Here we can show heat flowing from the surroundings to the system. Since energy is leaving the surroundings, the temperature of the water will decrease for an endothermic reaction. Since heat is being transferred to the system, we can go ahead and write heat on the reactant side, and Δh would be positive for an endothermic reaction.

So for an endothermic reaction, energy is transferred from the surroundings to the system, and therefore, the temperature of the water will decrease.