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

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. 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 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 25.35 °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) (74.65 °C) (x) = (100.0 g) (0.35 °C) (4.184 J g¯1 °C¯1)

Solving gives 0.131 J g¯1 °C¯1

Notice the rather small temperature gain by the water (25.00 to 25.35) and the very large temperature change (100 to 25.35) 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 farther 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.

Problem #3: a problem using data from an experiment

Problem #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

Problem #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.

Problem #6: Suppose a piece of iron with a mass of 21.5 g at a temp of 100.0 °C is dropped into an insulated container of water. The mass of the water is 132.0 g and its temperature before adding the iron is 20.0 °C. What will be the final temp of the system? Specific heat of iron is 0.449 kJ/kg K.

Solution:

1) Since

qlost, metal = qgained, water

we write

(mass) (Δt) (Cp, metal) = (mass) (Δt) (Cp, water)

2) Substituting:

(21.5) (100 - x) (0.449) = (132.0) (x - 20) (4.184)

Some explanation:

a) 100 - x is the Δt for the metal; it starts at 100.0 °C and drops to some unknown, final value.
b) x - 20 is the Δt for the water; it starts at 20.0 °C and rises to some unknown, final value.
c) Since both metal and water wind up at the same ending value, we need to use only one unknown for the two Δt expressions.

3) A wee bit of algebra:

(2150 - 21.5x) (0.449) = (132x - 2640) (4.184)

965.35 - 9.6535x = 552.288x - 11045.76

561.9415x = 12011.11

To 3 sig figs, the answer is 21.4 °C.

Problem #7: A 12.48 g sample of an unknown metal, heated to 99.0 °C was then plunged into 50.0 mL of 25.0 °C water. The temperature of the water rose to 28.1 °C. Assuming no loss of energy to the surroundings:

1. How many joules of energy did the water absorb?
2. How many joules of energy did the metal lose?
3. What is the heat capacity of the metal?
4. What is the specific heat of metal?

Solution:

1) q = (50.0 g) (3.1 °C) (4.184 J g-1 °C-1) = 648.52 J

2) 648.52 J

3) 648.52 J / 70.9 °C = 9.147 J/°C

4) 9.147 J/°C divided by 12.48 g = 0.733 J g-1 °C-1

Comment #1: this question doesn't use the qlost = qgained formulation of other questions. That is because the question is broken up into four parts. Notice that parts (1) and (2) are the equivalent of qlost = qgained and that (4) is the usual answer sought in problems of this type.

Comment #2: (3) is a step unnecessary to the solution for (4). It is there so you notice the difference between heat capacity and specific heat capacity.

Problem #8: A 43.2 g block of an unknown metal at 89.0 °C was dropped into an insulated vesssel containing 43.00 g of ice and 26.00 g of water at 0 °C. After the system had reached equilibrium it was determined that 9.15 g of the ice had melted. What is the specific heat of the metal? (The heat of fusion of ice = 334.166 J g¯1.)

Solution:

Comment: this variation of the usual suspects (detailed above) does NOT involve a temperature change in the water, only in the metal. Rather, some ice melts and the whole ice-water system stays at zero Celsius. Verrrry interesting!

1) Determine heat gained by the ice that melted:

9.15 g times 334.166 J g¯1 = 3057.62 J

2) Substitue and solve for the specific heat:

q = (mass) (Δt) (Cp, metal)

3057.62 J = (43.2 g) (89.0 °C) (x)

x = 0.795 J g¯1 °C¯1

Problem #9: A 35.0 g block of metal at 80.0 °C is added to a mixture of 100.0 g of water and 15.0 g of ice in an isolated container. All the ice melted and the temperature in the container rose to 10.0 °C. What is the specific heat of the metal?

Solution:

1) Determine heat required to melt the ice:

q = (15.0 g) (334.166 J g¯1) = 5012.49 J

Note that the 100 g of water is not mentioned yet.

2) Determine heat need to raise 115 g of water from 0 to 10.0 °C:

q = (115 g) (10.0 °C) (4.184 J g¯1 °C¯1) = 4811.6 J

Note the inclusion of the melted 15 g of ice. Also, notice that the water was at zero °C. We know this from the presence of the ice.

3) Determine the specific heat of the metal:

(5012.49 J + 4811.6 J) = (35.0 g) (70.0 °C) (x)

x = 4.01 J g¯1 °C¯1

Problem #10: A 500.0 g sample of an element at 153.0 °C is dropped into an ice-water mixture. 109.5 g of ice melts and there is still an ice - water mixture. What is the specific heat of the metal in J/g-°C? Given the molar heat capacity of the metal is 26.31 J/mol °C, what is the atomic weight and identity of the metal?

Solution:

1) Determine energy needed to melt the ice:

(6.02 kJ/mol) (109.5 g / 18.015 g/mol) = 36.5912 kJ

2) Determine specific heat:

36591.2 J = (500.0 g) (153.0 °C) (x)

x = 0.4783 J/g-°C

Note: we know the change in temperature is 153.0 °C because there is still ice in the water. That means the ice-water mix remained at zero Celsius as the 109.5 g of ice melted.

3) Determine aomic weight of the element:

0.4783 J/g-°C times x = 26.31 J/mol °C

x = 55.0 g/mol

The element is manganese.