Problem #2: How to Determine the Specific Heat of a Substance


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We are going to determine the specific heat of lead metal. Now this has already been done many times, so the value is available in reference books. We will pretend that is not the case.

Obviously, we need some pure lead, so we take a small piece of it. Let's say we use 49.51 grams. The shape does not matter.

We place the lead into an open beaker filled with boiling water and allow it to sit. We allow it to sit until all of the lead is the same temperature as the boiling water. We know what the temperature is, don't we?

It's 100.00 °C.

Now, how long it sat in the boiling water is immaterial, because we will assume it sat long enough. If you were doing this experiment for real, you might wind up doing the exact same experiment more than 100 times.

Now comes a real key step. As quickly as possible, we pull the metal out of the boiling water and transfer it into a beaker which holds 50.0 mL of some much cooler water, say 24.40 °C. We know this because we measured the temperature with a thermometer.

The hot lead cools down and the water heats up, until they both get to the same ending temperature. We record this with a thermometer and find that it is 27.20 °C. We now know two different Δt values. One is 100.00 minus the ending temperature (the lead) and the other is the ending temperature minus 24.40 (the water).

At this point we will make a key assumption which will make our task easier. That is to assume that all the heat lost by the lead winds up in the water. In reality this is not the case. In an actual experiment, the heat transfer will not be 100% and you have to take steps to compensate for those losses. We will ignore them.

The paragraph just above, when stated as a thermochemistry equation, is:

qlead = qwater

By substitution, we have (lead values on the left, water values on the right):

(mass) (Δt) (Cp) = (mass) (Δt) (Cp)

Putting the numbers in place gives us:

(49.51 g) (75.6 °C) (x) = (50.0 g) (2.8 °C) (4.184 J g¯1 °C¯1)

Solving gives 0.156 J g¯1 °C¯1

Notice the rather small temperature gain by the water (24.40 to 27.20) and the very large temperature change (100 to 27.20) of the lead. This is typical of problems of this sort.

Notice that 50.0 g of water is used in the calculation above, while farther above the text says 50.0 mL of water. The mass of water present is determined by volume times density. Since the density of water is 1.00 g mL¯1, the calculation is:

50.0 mL x 1.00 g mL¯1

with the answer being 50.0 g.

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