Thermochemistry Problems:
Two Equations Needed

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Example #1: How many kJ are required to heat 45.0 g of H2O at 25.0 °C and then boil it all away?

Solution:

Comment: We must do two calculations and then sum the answers.

1) The first calculation uses this equation:

q = (mass) (Δt) (Cp)

This summarizes the information needed:

Δt = 75.0 °C
The mass = 45.0 g
Cp = 4.184 J g¯1 °C¯1

2) Substituting, we have:

q = (45.0 g) (75.0 °C) (4.184 J g¯1 °C¯1)

q = 14121 J = 14.121 kJ

3) The second calculation uses this equation:

q = (moles of water) (ΔHvap)

This summarizes the information needed:

ΔHvap = 40.7 kJ/mol
The mass = 45.0 g
The molar mass of H2O = 18.0 gram/mol

4) Substituting, we obtain:

q = (45.0 g / 18.0 g mol¯1) (40.7 kJ/mol)

q = 101.75 kJ

5) Adding:

101.75 kJ + 14.121 kJ = 116 kJ (to three sig figs)

Example #2: How many kJ need to be removed from a 120.0 g sample of water, initially at 25.0 °C, in order to freeze it at 0 °C? (Area three, then area two on the time-temperature graph.)

Solution:

1) The first calculation:

q = (mass) (Δt) (Cp)

q = (120.0 g) (25.0 °C) (4.184 J g¯1 °C¯1)

q = 12,552 J = 12.552 kJ

2) The second calculation:

q = (moles of water) (ΔHvap)

q = (120.0 g / 18.0 g mol¯1) (6.02 kJ/mol)

q = 40.13 kJ

3) Summing up the values from the two steps gives 52.8 kJ (to three sig figs).


Example #3: You are given 12.0 g of ice at -5.00 °C. How much energy is needed to melt the ice completely to water?

Solution:

1) The first calculation:

q = (mass) (Δt) (Cp)

q = (12.0 g) (5.0 °C) (2.06 J g¯1 °C¯1)

q = 123.6 J = 0.1236 kJ

2) The second calculation:

q = (moles of water) (ΔHvap)

q = (12.0 g / 18.0 g mol¯1) (6.02 kJ/mol)

q = 4.0133 kJ

3) Summing up the values from the two steps gives 4.14 kJ, to three significant figures.


Example #4: Lead has a melting point of 327.5 °C, its specific heat is 0.128 J/g°C, and its molar enthalpy of fusion is 4.80 kJ/mol. How much heat, in kilojoules, will be required to heat a 500.0 g sample of lead from 23.0 °C to its melting point and then melt it?

Solution:

1) Two calculations are required:

1) heat iron from 23.0 to 327.5
2) melt iron at 327.5

2) Here are the calculation set-ups:

q1 = (500.0 g) (304.5 °C) (0.128 J/g°C) = 19488 J
q2 = (500.0 g / 55.845 g/mol) (4.80 kJ/mol) = 42.976 kJ

3) Add:

19.488 kJ + 42.976 kJ = 62.5 kJ (to three sig figs)

Note that I changed 19488 J to kJ before adding.


Example #5: The specific heat capacity of silver is 0.235 J/g-K. Its melting point is 962.0 °C, and its enthalpy of fusion is 11.3 kJ/mol. What quantity of energy, in Joules, is required to change 9.10 g of silver from a solid at 25.0 °C to a liquid at 962 °C?

Solution:

1) Two calculations are required:

1) heat silver from 25.0 to 962
2) melt silver at 962

2) Here are the calculation set-ups:

q1 = (9.10 g) (937.0 K) (0.235 J/g-K) = 2003.77 J
q2 = (9.10 g / 107.87 g/mol) (11.3 kJ/mol) = 0.953277 kJ = 953.277 J

3) The answer:

2003.77 J + 953.277 J = 2957.047 J

To three sig figs, 2960 J

Note how I use 937.0 K. This is because it is a difference, not an actual temperature. The difference between 962.0 °C and 25.0 °C is 937.0 K. You can see this by converting the two Celsius values to their Kelvin values and then subtracting. You'll get 937.0 K for the difference.


Example #6: What mass of CCl2F2 (heat of vaporization = 289 J/g) must evaporate in order to freeze 210. g of water initially at 24.0 °C? (The heat of fusion of water is 334 J/g; the specific heat of water is 4.184 J/g K)

Solution:

1) Let us determine how much heat is required to cool the water down and freeze it:

cool water from 24 to 0:
q1 = (210. g) (24 K) (4.184 J/g K)

q1 = 21087.36 J

freeze water:

q2 = (210. g) (334 J/g)

q2 = 70140 J

add 'em up:

q1 + q2 = 91227.36 J

2) Let us determine the mass of CCl2F2 required:

91227.36 J / 289 J/g = 316 g (to three sig figs)

Example #7: How much heat energy is required to convert 98.1 g of liquid sulfur dioxide, SO2, at 200. K to gaseous SO2 at 263 K? The molar heat of vaporization of SO2 is 24.9 kJ/mol, and the specific heat capacity of liquid SO2 is 1.36 J g¯1 °C¯1.

Solution:

1) In order to answer this question, we need to know the boiling point of SO2. Looking it up, we find 14 °C, which converts to 263 K.

2) Warm the liquid to its boiling point:

q = (mass) (Δt) (spec. heat)

q = (98.1 g) (63 °C) (1.36 J/g °C)

q = 8405.208 J

3) Vaporize the liquid:

q = (moles) (molar heat of fusion)

q = (98.1 g / 64.0638 g/mol) (24.9 kJ/mol) = 38.129 kJ

3) Add 'em up:

8.405208 kJ + 38.129 kJ = 46.5 kJ

Example #8: Calculate the amount of heat needed to melt 77.9 g of solid octane (C8H18) and bring it to a temperature of 119 °C. (Octane's molar heat of fusion is 21 kJ/mol. Its molar mass is 114.23 g/mol and the specific heat capacity for liquid octane is 255.68 J K¯1 mol¯1.)

Solution:

1) The first thing to do is assume the octane is as a solid and that it is at its melting point. Calculate the energy needed to melt the 77.9 g of solid octane at its melting point:

q = (moles) (molar heat of fusion)

q = (77.9 g / 114.23 g/mol) (21 kJ/mol)

q = 14.3211 kJ

2) The next step requires us to know the melting point and the boiling point of octane. So, we look those values up and find them to be −57.0 °C and 126 °C. That means our octane sample will fall short of boiling. All we need to do now is calculate the energy needed to warm octane from −57.0 °C to 119 °C:

q = (moles) (Δt) (spec. heat)

q = (77.9 g / 114.23 g/mol) (176 K) (255.68 J K¯1 mol¯1)

q = 30687.867 J

3) Add 'em up:

14.3211 kJ + 30.6879 kJ = 45.009 kJ

Two sig figs seems reasonable, so 45 kJ is the final answer.

4) Comments:

(a) Notice the unit of the specific heat in step 2, which caused me to calculate moles rather than use the mass.

(b) I also carried out a silent conversion from Celsius to Kelvin in step 2. Since the 176 is a difference, I can drop the °C and substitute K. This rests on the fact that the "size of 1 °C is equal to the "size" of 1 K.

(c) If you look up the boiling point and melting point of octane, you will find that it is given as a range, not as a single value. This is because octane is a mixture of a variety of C8H18 isomers. It's too expensive to get them all out so, depending on slight differences in the composition of the C8H18 sample you are testing, you will get a small variation in the melting and boiling points from sample to sample. I used a single value for the purposes of this question.


Example #9: A block of iron weighing 3.60 kg at a temperature of 807.0 °C was inserted into a container containing 1.00 liter of water at a temperature of 30.0 °C isolated from the environment. The cooling process of iron happens in two phases, the first phase the water is heated to the boiling point and in the second stage the water evaporates. This phase continues until the temperature of the iron is equal to the temperature of the water. The final temperature of the water and the iron is 100.0 °C. How many liters of liquid water will remain at the end of the process? (The specific heat of solid iron is 0.450 J g¯11 and the heat of vaporization for water is 2259.23 J/g.)

Solution:

1) We know what will happen to the iron: it will stay solid and will drop a total of 707 °C. How much heat energy is given off?

q = (3600 g) (707.0 K) (0.450 J g¯11) = 1145340 J

Note that I used 707 K rather than 707 °C. This is allowed, since this is a temperature difference and 1 K equals 1 °C in "size."

2) Let us determine how much heat is required to raise 1.00 L of water from 30.0 to 100.0 °C.

q = (1000 g) (70.0 °C) (4.184 J g¯1 °C¯1) = 292880 J

Note the silent change of 1.00 L to 1000 g.

3) How much energy is left?

1145340 − 292880 = 852460 J

4) Determine the amount of water that would be evaporated by 852460 J.

852460 J / 2259.23 J/g = 377.3 g

5) What volume of liquid water would remain?

1000 − 377.3 = 622.7 g

Using the density of water, this is 622.7 mL

to three sig figs, 0.623 L is the final answer.


Go to the Time-Temperature Graph file     Problems using four parts of the T-T graph
Problems using one part of the T-T graph     Problems using five parts of the T-T graph
Problems using three parts of the T-T graph     Return to Thermochemistry Menu