Thermochemistry Problems:One Equation Needed

The following problems mostly in area number 3 on the time-temperature graph. The most common type of question involves temperature changes, but the problems can be phrased to ask for a mass of substance involved or to ask you for the specific heat.

Less commonly are seen problems involving one of the other areas in the time-temperature graph. Some of the problems below are of this type.

Example #1: Calculate the heat necessary to raise 27.0 g of water from 10.0 °C to 90.0 °C

The important factor about this problem is that a ONLY temperature change is involved. Therefore, the equation to use is:

q = (mass) (Δt) (Cp)

This summarizes the information needed:

Δt = 80.0 °C
The mass = 27.0 g
Cp = 4.184 J g¯1 °C¯1

Only one calculation is needed and it is:

q = (27.0 g) (80.0 °C) (4.184 J g¯1 °C¯1)

q = 9037.44 J

Rounding to three significant figures and putting into kJ, we have 9.04 kJ for the final answer.

The usual practice is to convert Joule values above 1000 J to kJ (remember 1 kJ = 1000 J).

By the way, how did I know to use the specific heat value for liquid water? It is because I knew the melting point of water (0 ° C) and the boiling point of water (100 ° C). The temperature range was entirely within those two values, so I know that the water stayed as a liquid the entire time.

Less often, you will see problems which involve water changing temperature as a solid (area one on the time-temperature graph) or water changing temperature as a gas (area five on the time-temperature graph). Here is an example of each:

Example #2: Calculate the heat necessary to raise 27.0 g of water from −90.0 °C to −10.0 °C (graph area one)

Example #3: Calculate the heat necessary to raise 27.0 g of water from 110.0 °C to 190.0 °C (graph area five)

The technique is the same as in the liquid water example (area three of the graph) above. Here are the solutions:

The basic equation used in both solutions: q = (mass) (Δt) (Cp)

Example #2: q = (27.0 g) (80.0 °C) (2.06 J g¯1 °C¯1) = 4449.6 J (converted and rounded 4.45 kJ)

Example #3: q = (27.0 g) (80.0 °C) (2.02 J g¯1 °C¯1) = 4363.2 J (converted and rounded to 4.36 kJ)

Notice that all three of the above problems are exactly the same in mass and Δt. That was deliberate. Suppose the masses were 50.0 g. There would be no difference in technique, just that 50.0 g. would be the mass instead of 27.0 g. The same idea holds true for the change in temperature.

The critical thing to notice are the different specific heats. You MUST discern from the problem if the water (or whatever substance is in the problem) is a solid, liquid or gas. Then, you must select the proper specific heat (2.06 for solid, 4.184 for liquid or 2.02 for gas). Also, please note that your reference source may provide different values than the ones I am using.

To select the proper specific heat, you must know the melting and boiling temperature for the substance. Since water is the most-commonly discussed substance, I recommend you know the melting/freezing temperature for water (0 °C) and the boiling/condensing temperatue (100 °C) for water. Teachers (and textbook question writers) often assume you already know those two values.

What you would do is look at the temperatures in your problem. Do they fall below 0 (solid), between 0 to 100 (liquid) or above 100 (gas)? Also, if your starting or ending temperatures cross 0 or 100, then you know you have a problem that requires more than the one type of calculation discussed in this tutorial.

Please be aware that another type of "one equation required" problem might look like these two:

Example #4: How much energy is required to completely boil away 100.0 g of water at 100.0 °C? (area four on the graph)

Example #5: How much energy is required to melt 100.0 g of water at 0 °C? (area two on the graph)

The first problem requires the use of the molar heat of vaporization and the second requires the use of the molar heat of fusion.

Here are the two solutions:

40.7 kJ/mol x (100.0 g / 18.0 g/mol)

6.02 kJ/mol x (100.0 g / 18.0 g/mol)

Often these problems are solved using the heat of vaporization (2259 J/g) or the heat of fusion (334.166 J/g). These terms do not use the word molar. Here are the solutions:

2259 J/g x 100.0 g

334.166 J/g x 100.0 g

The heat of vaporization comes from 40700 J/mol divided by 18.015 g/mol and the heat of fusion comes from 6020 J/mol divided by 18.015 g/mol.

Example #6: Calculate the heat given off when 159.7 g of copper cools from 155.0 °C to 23.0 °C. The specific heat capacity of copper is 0.385 J/g °C.

Solution:

q = (159.7 g) (132.0 °C) (0.385 J/g °C)

q = 8115.954 J = 8.12 kJ (to three sig figs)

Example #7: How many joules of energy would be required to heat 15.9 g of diamond from 23.6 °C to 54.2 °C? (Specific heat capacity of diamond = 0.5091 J/g °C.)

Solution:

q = (15.9 g) (30.6 °C) (0.5091 J/g °C)

q = 247.697514 J = 248 J (to three sig figs)

This problem show a different twist to the "one equation" problems shown above:

Example #8: Assume that 491.8 J of heat is added to 5.00 g of water originally at 23.0 °C. What would be the final temperature of the water? (Specific heat capacity of water = 4.184 J/g °C)

Solution:

491.8 J = (5.00 g) (x) (4.184 J/g °C)

x = 23.5 °C

However, 23.5 is the change in temperature. To get the final temperature, we do this:

23.0 °C + 23.5 °C = 46.5 °C

Example #9: The temperature of a sample of water increases from 21.5 °C to 46.5 °C as it absorbs 5605 J of heat. What is the mass of the sample?

Solution:

q = (mass) (Δt) (Cp)

5605 J = (x) (25.0 °C) (4.184 J/g °C)

x = 53.6 g (to three sig figs)

Example #10: A 155 g sample of an unknown substance was heated from 25.0 °C to 40.0 °C. In the process, the substance absorbed 5696 J of energy. What is the specific heat of the substance?

Solution:

q = (mass) (Δt) (Cp)

5696 J = (155 g) (15.0 °C) (x)

x = 2.45 J/g °C

Note that the specific heat is being asked for.

Example #11: A 4.70 g nugget of pure gold absorbed 254 J of heat. What was the final temperature of the gold if the initial temperature was 22.0 °C? The specific heat of gold is 0.129 J/(g °C).

Solution:

q = (mass) (Δt) (Cp)

254 J = (4.70 g) (x) (0.129 J/(g °C))

x = 419 °C

However, this is the change in temperature, not the final temperature. The final temp is:

22 + 419 = 441 °C

Example #12: An engineer is working on a new engine design. One of the moving parts is an alloy that contains 1.60 kg of aluminum and 0.300 kg of iron and is designed to operate at 210 °C. How much heat is required to raise its temperature from 20.0 °C to 210. °C?

Solution:

We will do two separate energy calculations and add them together.

1) First, the aluminum:

q = (1.60 kg) (190 K) (0.900 kJ/(kg K))

q = 273.6 kJ

2) Next, the iron:

q = (0.300 kg) (190 K) (0.450 kJ/(kg K))

q = 25.65 kJ

273.6 + 25.65 = 299 kJ (to three sig figs)

Note unit for specific heat. Notice that it is numerically equal to the more common unit of J/(g °C).

The temperature difference between 20 °C and 210 °C is 190 °C. Since it is a difference, it can also be expressed as 190 K. This is because the "size" of 1 °C is equal to the "size" of 1 K.

If you are still not convinced, convert 20 and 210 to their respective Kelvin values and subtract. The answer will be 190 K.

Example #13: What is the final temperature if 9.00 grams of metal (specific heat = 0.900 J/g °C) at 58.0 °C receives 292.0 J of heat from the surroundings?

Solution:

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

292 J = (9.00 g) (x) (0.900 J/g °C)

x = 36.049 °C

However, this is the total change, we need the final temperature:

58.0 + 36.049 = 94.049 °C

To three significant figures, this is 94.0 °C

Example #14: Water at 78.0 °C is cooled to 15.0 °C and is found to release 1.12 x 105 J of energy. What is the mass of the water?

Solution:

q = (mass) (temperature change) (specific heat)

112000 J = (x) (63.0 °C) (4.184 J g¯1 °C¯1)

x = 425 g

Example #15: 1.00 kg of liquid mercury was heated from −20.0 °C to 25.0 °C. Find the heat energy required. (specific heat of liquid Hg = 0.138 J/g °C).

Solution:

q = (mass) (Δt) (Cp)

q = (1000 g) (45.0 °C) (0.138 J/g °C) <--- note change from kg to g

q = 6210 J

Example #16: A 55.0 g piece of metal at 95.0 °C is cooled down to 30.0 °C and releases 1.63 x 103 J of energy. Calculate the specific heat of the metal.

Solution:

q = (m) (Δt) (Cp)

1630 J = (55.0 g) (65.0 °C) (x)

x = 0.456 J g¯1 °C¯1

Example #17: 350. g of water at 45.0 °C released 25.0 kJ of heat. Find the new temperature.

Solution:

25000 J = (350. g) (x) (4.184 J g¯1 °C¯1) <--- note change from kJ to J

x = 17.1 °C <--- this is the temp change, not the final answer

final temp ---> 45.0 − 17.1 = 27.9 °C

Note: if the problm had said absorbed 25.0 kJ, we would have added 17.1 to 45.0 for the final answer.

Example #18: Calculate the amount of heat required to raise the temperature of a 55.0 g sample of water from 34.0 °C to 69.0 °C.

Solution:

q = (55.0 g) (35.0 °C) (4.184 J g¯1 °C¯1)

q = 8054.2 J

to three sig figs, 8050 J