### How to Determine the Specific Heat of a Substance

Example #1: We are going to determine the specific heat of copper 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 copper, so we take a small piece of it. Let's say we use 15.0 grams. The shape does not matter.

We place the copper metal into an open beaker filled with boiling water and allow it to sit. We allow it to sit until all of the copper metal 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.

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 100.0 mL of some much cooler water, say 25.00 °C. We know this because we measured the temperature with a thermometer.

The hot copper metal 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 26.02 °C. We now know two different Δt values. One is 100.00 minus the ending temperature (the copper) and the other is the ending temperature minus 25.00 (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 copper 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:

qcopper = qwater

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

(mass) (Δt) (Cp) = (mass) (Δt) (Cp)

Putting the numbers in place gives us:

(15.0 g) (73.98 °C) (x) = (100.0 g) (1.02 °C) (4.184 J g¯1 °C¯1)

Solving gives 0.384 J g¯1 °C¯1

Notice the rather small temperature gain by the water (25.00 to 26.02) and the large (by comparison) temperature change (100 to 26.02) of the copper. This is typical of problems of this sort.

Notice that 100.0 g of water is used in the calculation above, while further above the text says 100.0 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:

100.0 mL x 1.00 g mL¯1

with the answer being 100.0 g.

Example #2: The exact same text as above, just replacing copper with lead and using different numbers.

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.

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:

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) (72.8 °C) (x) = (50.0 g) (2.8 °C) (4.184 J g¯1 °C¯1)

Solving gives 0.1625 J g¯1 °C¯1. Following the rule for rounding off with five, the final answer is 0.162 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 further 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.

Example #3: We are going to determine the specific heat of a metal using experimental data. In this experiment, we used a "coffee cup" calorimeter and gathered the following data:

 Mass of empty cup 2.31 g Mass of cup + water 180.89 g Mass of cup + water + metal 780.89 g Initial temperature of water 17.0 °C Initial temperature of metal 52.0 °C Final temperature of system 27.0 °C

The key thermochemistry equation for solving this problem is:

qmetal = qwater

Then, by substitution, we have (metal values on the left, water values on the right):

(mass) (Δt) (Cp) = (mass) (Δt) (Cp)

We need to work with values from the data table to get what we need to substitute into the above equation.

mass of water: 180.98 − 2.31 = 178.58 g

mass of metal: 780.89 − 180.89 = 600.0 g

change in water temperature: 27.0 − 17.0 = 10.0 °C

change in metal temperature: 52.0 − 17.0 = 25.0 °C

Putting the numbers in place gives us:

(600.0 g) (25.0 °C) (x) = (178.58 g) (10.0 °C) (4.184 J g¯1 °C¯1)

Solving gives 0.498 J g¯1 °C¯1

Notice the starting temperature of the metal (52.0 °C). This is an unusual value in that the metal sample is usually heated up by immersion in boiling water, making the usual starting temperature at or near 100.0 °C for the metal.

Often, problems of this sort will specify mL of water rather than grams. 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:

your mL x 1.00 g mL¯1

with the answer being the same numerical value, just with grams as the unit rather than mL.

Example #4: A piece of metal weighing 59.047 g was heated to 100.0 °C and then put it into 100.0 mL of water (initially at 23.7 °C). The metal and water were allowed to come to an equilibrium temperature, determined to be 27.8 °C. Assuming no heat lost to the environment, calculate the specific heat of the metal.

qmetal = qwater

(mass) (Δt) (Cp) = (mass) (Δt) (Cp)

(59.047 g) (72.2 °C) (x) = (100.0 g) (4.1 °C) (4.184 J g¯1 °C¯1)

x = 0.402 J g¯1 °C¯1

Example #5: A 25.6 g piece of metal was taken from a beaker of boiling water at 100.0 °C and placed directly into a calorimeter holding 100.0 mL of water at 25.0 °C. The calorimeter heat capacity is 1.23 J/K. Given that the final temperature at thermal equilibrium is 26.2 °C, determine the specific heat capacity of the metal.

Solution:

1) We know this:

qlost, metal = qgained

2) However, energy is gained by two different entities (the water and the calorimeter itself). Therefore:

qlost, metal = qgained, water + qgained, calorimeter

3) Substituting, we have:

(mass) (Δt) (Cp, metal) = (mass) (Δt) (Cp, water) + (Δt of water) (calorimeter constant)

4) Putting values into place and solving:

(25.6 g) (73.8 °C) (x) = (100.0 g) (1.2 °C) (4.184 J/g °C) + (1.2 °C) (1.23 J/K)

x = 0.266 J/g °C

Comment #1: the °C and the K cancel in this case because (1) one °C is the same size as one K and (2) the 1.2 °C is a temperature difference, not a temperature of 1.2 °C.

Comment #2: we could make a tenative identification of the metal as niobium, based on its specific heat. See here.

Example #6: When 12.29 g of finely divided brass at 95.0 °C is quickly stirred into 40.00 g of water at 22.0 °C in a calorimeter, the water temperature rises to 24.0 °C. Find the specific heat of brass.

Solution:

1) Let us use the following specific heat of water:

4186 J kg¯11

2) Determine the energy to heat the water:

q = (mass) (change in temp) (specific heat)

q = (0.04000 kg) (2.0 K) (4186 J kg¯11) = 334.88 J

3) The energy lost by the brass as it cooled is the same amount absorbed by the water:

q = (mass) (change in temp) (specific heat)

334.88 J = (0.01229 kg) (71.0 K) (x)

x = 384 J kg¯11

or, if you prefer, 0.384 J g¯1 °C¯1

Example #7: When 450. g of metal shot at 100.0 °C is quickly poured into a hole in a block of ice at 0.00 °C, 25.0 g of ice melts.What is the specific heat of the metal?

Solution:

Since ice remains, the liquid water stays at 0.00 °C.

(25.0 g) (334.166 J/g) = 8354.15 J (amount of heat lost by metal shot)

q = (mass) (Δt) (specific heat)

8354.15 J = (450. g) (100.0 °C) (Cp)

Cp = 0.186 J g¯1 °C¯1 (to three sig figs)