Thermochemistry Example Problems:
Four or Five Equations Needed


Some examples of problems covering one area of the time temperature graph.

Some examples of problems covering two or three areas of the time temperature graph.

Go to the Time-Temperature Graph tutorial

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Four Equations Needed

Problem #1: 33.3 grams of ice at 0.00 °C has heat added to it until steam at 150.0 °C results. Calculate the total energy expended. (Hint: melt, raise, boil, raise.)

Go to answer.


Problem #2: How much heat is released (in kJ) when 105.0 g of steam at 100.0 °C is cooled to ice at -15.0 °C? (Hint: condense, cool, freeze, cool) Please use these values:

Enthalpy of vaporization of water = 40.67 kJ/mol
Enthalpy of fusion for water = 6.01 kJ/mol
molar heat capacity of liquid water = 75.4 J mol¯1 °C¯1
molar heat capacity of ice = 36.4 J mol¯1 °C¯1

Comment: note the use of mol in the specific heat values as opposed to the more common use of grams.

Solution:

1) Condense 105.0 g of water at 100.0 °C:

(40.67 kJ/mol) (105.0 g / 18.015 g/mol) = 237.044 kJ

2) Cool the liquid water from 100 °C to 0 °C:

(105.0 g / 18.015 g/mol) (100.0 °C) (75.4 J mol¯1 °C¯1) = 43946.7 J

3) Freeze the water at 0 °C:

(6.01 kJ/mol) (105.0 g / 18.015 g/mol = 35.029 kJ

4) Cool the ice from 0 to -15 °C:

(105.0 g / 18.015 g/mol) (15.0 °C) (36.4 J mol¯1 °C¯1) = 3182.348 J

5) Add the above four results:

237.044 kJ + 43.9467 kJ + 35.029 kJ + 3.182348 kJ = 319.2 kJ

to three sig figs, the answer is 319 kJ


Problem #3: How much heat is required to completely vaporize 4.80 g of ice which is at -30.0 °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

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 the calculations that use specific heats are temperature differences, not an absolute temperature value.

Solution:

1) raise 4.80 g of ice from -30.0 to zero Celsius:

(4.80 g) (30.0 K) (2.06 J g¯11) = 296.64 J

2) melt 4.80 g of ice:

(4.80 g) (334.16 J g¯1) = 1603.968 J

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

(4.80 g) (100.0 K) (4.184 J g¯11) = 2008.32 J

4) evaporate 4.80 g of liquid:

(4.80 g) (2259 J g¯1) = 10843.2 J

5) add the results:

296.64 + 1603.968 + 2008.32 + 10843.2 = 156173 J = 14752.128 = 14.75 kJ

Five Equations Needed

Problem #4: 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

Problem #5: 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 moles of water (207.0 / 18.015 = 11.49).

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 sf, this would be 638 kJ.


Problem #6: 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.


Problem #7: 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.

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)

Problem #8: 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)

Some examples of problems covering one area of the time temperature graph.

Some examples of problems covering two or three areas of the time temperature graph.

Go to the Time-Temperature Graph file

Return to Thermochemistry Menu