Transcript for:
Understanding Calorimetry Techniques

Now that we are aware of some basic calorimetry setups, what are some experiments we could perform to gather thermochemical data? Something that can be done with a simple calorimeter would be to measure the heat capacity of an unknown object. A piece of metal could be heated to a known temperature, perhaps in boiling water, and then transferred to a known volume of room temperature water in a calorimeter. By measuring the temperature change of the water, as the hot metal transfers heat into the water, we can measure the amount of heat transferred into the water since we also know the mass of the water given its volume as well as the specific heat of water. Specific heat, mass, and change in temperature is everything we need to solve for Q in this equation for the water. We also know from conservation of energy that there can have been no loss of heat during this transfer. In other words, Q water the heat absorbed by the water, plus Q metal, the heat released by the metal, equals zero. If we subtract Q metal from both sides, we get Q water equals negative Q metal, which represents the reasonable assumption that all of the heat involved in the process went directly from the metal to the water, losing only a negligible amount to the environment. Therefore, in calculating the heat absorbed by the water in the calorimeter, We will also know the amount of heat that was released from the metal, which is simply the negative version of Q for the water. With this information, we can plug values into a new equation, but this time using values for the metal. The Q value will be the Q metal we just got from our calculations for water, but this time the mass will be the mass of the metal, and the change in temperature will go from the initial temperature of the hot metal to the final temperature in the calorimeter. The delta T will therefore be negative this time, which makes sense as it will cancel out the negative in the Q value. The only unknown is the specific heat, so we can solve for it. Once we have the specific heat, we can hopefully identify the composition of the unknown metal. For example, let's say a 59.7 gram piece of metal submerged in boiling water is transferred into 60 milliliters of water initially at 22 degrees Celsius. The temperature of the water is then raised to 28.5 degrees. What is the specific heat of the metal? Just as we discussed, let's first use the given data to figure out the heat absorbed by the water using this equation. Since water is one gram per milliliter, it's easy to get the mass of the water. The specific heat of water is a known value that is always the same, and we know the initial and final temperature of the water. So let's plug those values into the equation. We should get 1,632 joules, which is the heat absorbed by the water. Let's also recall that the heat absorbed by the water must be the heat released by the metal. So Q metal must be negative 1,632 joules. since a positive Q means heat absorbed and a negative Q means heat released. So let's write a new equation using the Q metal we just calculated. This time, we have the mass of the unknown metal and the temperature change experienced by the metal, and the only unknown will be the specific heat of the metal. Doing the math, we should get 0.382 joules per gram Celsius. as the specific heat of the metal. If we take a look at our list of specific heats, we might notice that the tabulated specific heat for copper is extremely close to the value we just calculated for our unknown metal, making it very likely that this was a piece of copper.