Thermochemistry Examples:
Five Equations Needed

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Example #1: Calculate the amount of energy required to change 50.0 g of ice at -20.0 °C to steam at 135.0 °C. Please use these values:

Heat of fusion = 334.16 J g¯1
Heat of vaporization = 2259 J g¯1
specific heat capacity for solid water (ice) = 2.06 J g¯11
specific heat capacity for liquid water = 4.184 J g¯11
specific heat capacity for gaseous water (steam) = 2.02 J g¯11

Solution:

1) raise 50.0 g of ice from -20.0 to zero Celsius:

(50.0 g) (20.0 K) (2.06 J g¯11) = 2060 J

2) melt 50.0 g of ice:

(50.0 g) (334.16 J g¯1) = 16708 J

3) raise 50.0 g of liquid water from zero to 100.0 Celsius:

(50.0 g) (100.0 K) (4.184 J g¯11) = 20920 J

4) evaporate 50.0 g of liquid:

(50.0 g) (2259 J g¯1) = 112950 J

5) raise 50.0 g of steam from 100.0 to 135.0 Celsius:

(50.0 g) (35.0 K) (2.02 J g¯11) = 3535 J

6) add the results:

2060 + 16708 + 20920 + 112950 + 3535 = 156173 J = 156 kJ

Example #2: Calculate the amount of energy in kilojoules needed to change 207.0 g of water ice at -10.0 °C to steam at 125.0 °C. The following constants for water may be helpful.

Cp, ice = 36.39 J mol¯1 °C¯1
Cp, liquid = 75.375 J mol¯1 °C¯1
Cp, steam = 37.11 J mol¯1 °C¯1
ΔHfus = 6.02 kJ mol¯1
ΔHvap = 40.7 kJ mol¯1

Comment: notice the unit on the specific heat values. It uses 'per mole' rather than 'per gram.' This means the either (1) we have to change the specific heat values to the 'per gram' value (do this by dividing by the molar mass of water) or (2) converting the grams of water to moles of water (by dividing by the molar mass of water).

For this example, let us use 11.49 moles of water (from 207.0 g / 18.015 g/mol).

Solution:

1) raise 11.49 moles of ice from -10.0 to zero Celsius:

(11.49 moles) (10.0 °C) (36.39 J mol¯1 °C¯1) = 4181 J

2) melt 11.49 moles of ice:

(11.49 moles) (6.02 kJ mol¯1) = 69.170 kJ = 69,170 J

3) raise 11.49 moles of liquid water from zero to 100.0 Celsius:

(11.49 moles) (100.0 °C) (75.375 J mol¯1 °C¯1) = 86,606 J

4) evaporate 11.49 moles of ice:

(11.49 moles) (40.7 kJ mol¯1) = 467.643 kJ = 467,643 J

5) raise 11.49 moles of steam from 100.0 to 125.0 Celsius:

(11.49 moles) (25.0 °C) (37.11 J mol¯1 °C¯1) = 10,660 J

6) add the results:

4181 + 69,170 + 86,606 + 467,643 + 10,660 = 638260 J = 638.26 kJ

To three sig figs, this would be 638 kJ.


Example #3: If 53.2 kJ of heat are added to a 15.5 g ice cube at -5.00 °C, what will be the resulting state and temperature of the water?

Solution:

1) Determine kJ needed to heat ice from -5 °C to zero °C:

q = (15.5 g) (5.00 °C) (2.06 J g¯1 °C¯1) = 159.65 J = 0.15965 kJ

2) Determine kJ needed to melt the ice:

q = (15.5 g / 18.015 g mol¯1) (6.02 kJ mol¯1) = 5.1796 kJ

3) Determine kJ needed to heat water from zero °C to 100 °C:

q = (15.5 g) (100.0 °C) (4.184 J g¯1 °C¯1) = 6485.2 J = 6.4852 kJ

4) Determine kJ needed to vaporize the liquid water:

q = (15.5 g / 18.015 g mol¯1) (40.7 kJ mol¯1) = 35.018 kJ

5) Let us determine how many kJ expended to this point:

0.15965 + 5.1796 + 6.4852 + 35.018 = 46.84245 kJ

6) How many kJ remain?

53.2 - 46.84245 = 6.35755 kJ

7) Determine the temperature change that 6.35755 kJ induces in 15.5 g of steam:

6357.55 J = (15.5 g) (x) (2.02 J g¯1 °C¯1) = 203 °C

Since the steam started at 100 °C, the final temperature of the steam is 303 °C.

Comment: In a problem like this, you could maintain a running total, if so desired. Also, watch out for a time when your total (if you don't keep a running total) exceeds the kJ available in the problem. You'll have to back up, get a new subtotal and figure out how many kJ are left to be consumed in the last step.


Example #4: Calculate the heat required to convert 15.4 g of ethyl alcohol, C2H5OH, from a solid at -131.0 °C into the gaseous state at 104.0 °C. The normal melting and boiling points of this substance are -117 °C and 78 °C, respectively. The heat of fusion is 109 J/g, and the heat of vaporization is 837 J/g. The specific heats of the solid, liquid and gaseous states are, respectively, 0.97, 2.30 and 0.95 J/g-K.

Comment: please note the use of J/g values as opposed to kJ/mol. Also note the use of K in the specific heat capacities. This use of K does not affect the calculations because (1) the size of one K is the same as the size of one C and (2) the temperature values in #1, 3, and 5 below are temperature differences, not an absolute temperature value.

Warning: sometimes, a teacher will teach using °C in the specific heat and then provide constants on the test that utilize K (instead of °C). Beware!

Solution:

1) raise 15.4 g of solid from -131.0 to -117.0 Celsius:

(15.4 g) (14.0 K) (0.97 J g¯11) = 209.132 J

2) melt 15.4 g of solid:

(15.4 g) (109 J g¯1) = 1678.6 J

3) raise 15.4 g of liquid alcohol from -117.0 to 78.0 Celsius:

(15.4 g) (195.0 K) (2.30 J g¯11) = 6906.9 J

4) evaporate 15.4 g of liquid:

(15.4 g) (837 J g¯1) = 12889.8 J

5) raise 15.4 g of gaseous alcohol from 78.0 to 104.0 Celsius:

(15.4 g) (26.0 K) (0.95 J g¯11) = 380.38 J

6) add the results:

209.132 + 1678.6 + 6906.9 + 12889.8 + 380.38 = 22064.812 J = 22.1 kJ (to three sig fig)

Comment: notice how I used K for the temperature changes. That's because the temperatures are temperature differences, not specific temperature values. The 195.0 K is the difference between -117 °C and 78 °C. You can convert the two °C values to K by adding 273 to each. The difference between those two K values is 195.0 K.


Example #5: 15.0 g of a gas starting at 120.0 °C is cooled to a solid at 0.0 °C. The substance undergoes phase changes at 85.0 °C and 10.0 °C. The heat of condensation is 3.00 kJ/g and the heat of crystallization is 1.00 kJ/g. The heat capacity of the gas is 0.100 kJ/g °C. The heat capacity of the liquid is 0.0500 kJ/g °C. The heat capacity of the solid is 0.0300 kJ/g °C.

Determine the following:

a) q of gas cooling
b) q of condensation
c) q of liquid cooling
d) q of crystallization
e) q of solid cooling
f) total heat of process (start to finish)

Solution:

1) q of gas cooling:

q = (15.0 g) (35.0 °C) (0.100 kJ/g °C) = 52.5 kJ

2) q of condensation:

q = (15.0 g) (3.00 kJ/g) = 45.0 kJ

3) q of liquid cooling:

q = (15.0 g) (75.0 °C) (0.0500 kJ/g) = 56.25 kJ

4) q of crystallization:

q = (15.0 g) (1.00 kJ/g) = 15.0 kJ

5) q of solid cooling:

q = (15.0 g) (10.0 °C) (0.0300 kJ/g) = 4.50 kJ

6) total heat of process:

add the results: 52.5 + 45.0 + 56.25 + 15.0 + 4.50 = 173.25 kJ (best answer to report would be 173.2 kJ)

Example #6: Calculate the amount of energy in kilojoules needed to change 117.0 g of water ice at -10.00 °C to steam at 125.0 °C. The following constants will be used:

Cp (ice) = 36.57 J/(mol °C)
Cp (water) = 75.40 J/(mol °C)
Cp (steam) = 36.04 J/(mol °C)
ΔHfus = 6.02 kJ/mol
ΔHvap = 40.67 kJ/mol

Solution:

1) Since all the constants utilize moles, let us convert 117.0 g of water to moles:

117.0 g / 18.015 g/mol = 6.49459 mol <--- I'll carry a couple guard digits

2) Five calculations will be used because the water does five different behaviors:

a) warm up (as a solid) from -10 °C to zero.
b) while at zero °C, melt
c) warm up (as a liquid) from zero °C to 100.
d) while at 100 °C, boil
e) warm up (as a gas) from 100 to 125 °C

3) The five calculations:

a) q = (6.49459 mol) (10.00 °C) (36.57 J/(mol °C)) = 2375.07 J
b) q = (6.49459 mol) (6.02 kJ/mol) = 39.09743 kJ
c) q = (6.49459 mol) (100.0 °C) (75.40 J/(mol °C)) = 48969.21 J
d) q = (6.49459 mol) (40.67 kJ/mol) = 264.135 kJ
e) q = (6.49459 mol) (25.0 °C) (36.04 J/(mol °C)) = 5851.6256 J

4) A final summing up (note conversions of J to kJ):

2.37507 + 39.09743 + 48.96921 + 264.135 + 5.8516256 = 360.428 kJ

To four significant figures, the answer is 360.4 kJ


Example #7: How much energy is required to heat 41.0 g of H2O(s) at -24.0 °C to H2O(g) at 147.0 °C? Some constants follow:

Enthalpy of fusion333.6 J/g6010. J/mol
Enthalpy of vaporization2257 J/g40660 J/mol
Specific heat of solid H2O (ice)2.087 J/(g °C)37.60 J/(mol °C)
Specific heat of liquid H2O (water)4.184 J/(g °C)75.37 J/(mol °C)
Specific heat of gaseous H2O (steam)2.000 J/(g °C)36.03 J/(mol °C)

Comment: notice that both gram-based and mole-based units are given. Let's do solutions using both sets. I won't mix-and-match!

Solution using grams:

1) A summary of the five behaviors we will set up calculations for:

q1 ---> heat solid water from -24 to 0
q2 ---> melt solid water at 0
q3 ---> heat liquid water from 0 to 100
q4 ---> boil liquid water at 100
q5 ---> heat up gaseous water from 100 to 147

2) The five calculations:

q1 = (41.0 g) (24.0 °C) (2.087 J/(g °C)) = 2053.608 J
q2 = (41.0 g) (333.6 J/g) = 13677.6 J
q3 = (41.0 g) (100. °C) (4.184 J/(g °C)) = 17154.4 J
q4 = (41.0 g) (2257 J/g) = 92537 J
q5 = (41.0 g) (47.0 °C) (2.000 J/(g °C)) = 3854 J

3) The sum of the five q values:

129276.608 J

Rounded to three significant figures and converted to kJ:

129 kJ

Solution using moles:

1) Convert grams to moles:

41.0 g / 18.015 g/mol = 2.27588121 mol

2) The five calculations:

q1 = (2.27588121 mol) (24.0 °C) (37.60 J/(mol °C)) = 2053.7552 J
q2 = (2.27588121 mol) (6010 J/mol) = 13678.046 J
q3 = (2.27588121 mol) (100. °C) (75.37 J/(mol °C)) = 17153.3 J
q4 = (2.27588121 mol) (40660 J/mol) = 92537.33 J
q5 = (2.27588121 mol) (47.0 °C) (36.03 J/(mol °C)) = 3854 J

3) Add them together and convert to kJ:

129276.4312 J = 129 kJ (to three sig figs)

Note: you could have used (41.0 g / 18.015 g/mol) in place of (2.27588121 mol), if so desired. You would just have to push a few extra buttions to enter the division each time. Purely a stylistic decision as to which way to do it.


Example #8: How much energy is required to turn 18.0 g of ice at 228 K into steam at 418 K? Use these constants:

Specific heat of ice = 2.077 J/g K
Specific heat of water (ℓ) = 4.184 J/g K
Specific heat of steam = 2.042 J/g K
H2O heat of fusion = 6.02 kJ/mol
H2O heat of vaporization = 40.67 kJ/mol

Solution:

1) A summary of what the water does:

q1 ---> heats up from 228 K to 273 K (ice)
q2 ---> melts at 273 K
q3 ---> heats up from 273 to 373 K (liquid)
q4 ---> boils at 373 K
q5 ---> heats up from 373 to 418 K (steam)

2) The equations:

q1 ---> (18.0 g) (45 K) (2.077 J/g K)
q2 ---> (18.0 g / 18.0 g/mol) (6.02 kJ/mol)
q3 ---> (18.0 g) (100 K) (4.184 J/g K)
q4 ---> (18.0 g / 18.0 g/mol) (40.67 kJ/mol)
q5 ---> (18.0 g) (45 K) (2.042 J/g K)

3) The energy from each step:

q1 ---> 1682.37 J
q2 ---> 6.02 kJ
q3 ---> 7531.2 J
q4 ---> 40.67 kJ
q5 ---> 1654.02 J

4) Add 'em up:

57.55759 kJ

57.6 kJ (to three sig figs, note use of the 'rounding off a five' rule)

Remember to convert q1, q3, q5 to kJ before adding.


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