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# Constant-pressure calorimetry

AP.Chem:
ENE‑2 (EU)
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ENE‑2.D (LO)
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ENE‑2.D.1 (EK)
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ENE‑2.D.2 (EK)
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ENE‑2.D.3 (EK)
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ENE‑2.D.4 (EK)
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ENE‑2.D.5 (EK)
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ENE‑2.D.6 (EK)

## Video transcript

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 and 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 and therefore the temperature of the water will increase which we will we will see on the thermometer so 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 and 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 is equal to mc delta t 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 delta t is the change in the temperature which would be the final temperature so 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 so the grams cancel out uh degrees celsius cancels out and we're left with joules as our unit so q is equal to uh 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 .39 joules per gram degree celsius and 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 uh grams will cancel degrees celsius will cancel and our answer will be in joules so q is equal to using two significant figures negative three point negative three point three times ten to the third joules and the negative sign so this 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 all right we can see that we've lost more 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 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 so before we do that let's review some terms for thermodynamics so 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 are ever everything else which would include the water in the calorimeter the coffee cup itself the thermometer the environment outside so the surroundings are everything else and finally the universe would be the system plus the surroundings so the reactants and the products make up the system so that's what the s stands for here in in our calorimeter that's our system let's say we run a reaction and in the reaction heat is given off so in that case heat would flow from the system to the surroundings and so the temperature of the water would increase so 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 is equal to m c delta t equation and 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 sine next remember that our lid over here is loose fitting which makes this constant pressure calorimetry and 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 delta h so we can write that qp is equal to delta h and when delta h is negative we're talking about an exothermic an exothermic reaction so 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 so 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 and since heat is being transferred to the system we can go ahead and write heat over here we can go ahead and write heat on the reactant side and delta h 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
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