Thermochemistry Example Problems:
Two or Three Equations Needed

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

Some examples of problems covering four or five areas of the time temperature graph.

Go to the Time-Temperature Graph tutorial

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Two Equations Needed

Problem #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)

Problem #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.


Problem #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.


Problem #4: Equal masses of hot water and ice are mixed together. All of the ice melts and the final temperature of the mixture is 0 °C. If the ice was originally at 0 °C, what was the initial temperature of the hot water?

Solution:

Let us assume we have 18.0 g of ice and 18.0 g of hot water present.

The key is to realize that the only thing the ice did is melt, it did not change its temperature. So, let us calculate the amount of heat needed to melt our 18.0 g (or, 1.00 mole) of ice:

q = (6.02 kJ/mol) (1.00 mol) = 6.02 kJ

The only source of heat is the hot water, which provides 6020 J (I converted the 6.02 kJ to J.) of heat. Let us calculate the temperature change of 18.0 g of hot water as it loses 6020 J of heat:

6020 J = (18.0 g) (x) (4.184 J g¯1 °C¯1)

x = 79.9 °C

The hot water was at an initial temperature of 79.9 °C (since everything ended up at a final temperature of 0 °C.


Three Equations Needed

Problem #5: A 36.0 g sample of water is initially at 10.0 °C. How much energy is required to turn it into steam at 200.0 °C? (This example starts with a temperature change, then a phase change followed by another temperature change.)

Solution:

q = (36.0 g) (90.0 °C) (4.184 J g¯1 °C¯1) = 13,556 J = 13.556 kJ

q = (40.7 kJ/mol) (36.0 g / 18.0 g/mol) = 81.4 kJ

q = (36.0 g) (100.0 °C) (2.02 J g¯1 °C¯1) = 7272 J = 7.272 kJ

q = 102 kJ (rounded to the appropriate number of significant figures)


Problem #6: A 72.0 g sample of ice is at 0 °C. How much energy is required to convert it to steam at 100.0 °C? (This example begins with phase change, then a temperature change and then a second phase change, areas two, three and four on the time-temperature graph.)

Solution:

q = (6.02 kJ/mol) (72.0 g / 18.0 g/mol) = 24.08 kJ

q = (72.0 g) (100.0 °C) (4.184 J g¯1 °C¯1) = 30,125 J = 30.125 kJ

q = (40.7 kJ/mol) (72.0 g / 18.0 g/mol) = 162.8 kJ

q = 217 kJ (rounded to the appropriate number of significant figures)


Problem #7: Calculate the heat released by cooling 54.0 g H2O from 57.0 °C to minus 3.0 °C. (Cools from 57.0 to zero, a phase change, the cools from zero to -3; three equations needed.)

Solution:

q = (54.0 g) (57.0 °C) (4.184 J g¯1 °C¯1) = 12,878 J = 12.878 kJ

q = (6.02 kJ/mol) (54.0 g / 18.0 g/mol) = 18.06 kJ

q = (54.0 g) (3.0 °C) (2.06 J g¯1 °C¯1) = 333.73 J = 0.334 kJ

q = 31.3 kJ (rounded to the appropriate number of significant figures)


Problem #8: How much energy is required to heat 125.0 g of water from -15.0 °C to 35.0 °C?

The solution:

q = (125.0 g) (15.0 °C) (2.06 J g¯1 °C¯1) = 3862.5 J = 3.8625 kJ

q = (6.02 kJ/mol) (125.0 g / 18.0 g/mol) = 41.806 kJ

q = (125.0 g) (35.0 °C) (4.184 J g¯1 °C¯1) = 18,305 J = 18.305 kJ

q = 64.0 kJ (rounded to the appropriate number of significant figures)


Problem #9: 18.0 mL of water at 28.0 °C are added to a hot skillet. All of the water is converted to steam at 100.0 °C. The mass of the pan is 1.25 kg and the molar heat capacity of iron is 25.19 J/mol °C. What is the temperature change of the skillet?

Solution:

1) Determine the energy need to heat and boil the water:

heat: q = (18.0 g) (72.0 °C) (4.184 J/g °C) = 5422.464 J = 5.422464 kJ

boil: q = (40.7 kJ/mol) (18.0 g / 18.0 g/mol) = 40.7 kJ

total: 40.7 kJ + 5.422464 kJ = 46.122464 kJ

I won't bother to round off until the final answer.

2) Determine temperature change of skillet:

46122.464 J = (22.38338 mol) (Δt) (25.19 J/mol °C)

Δt = 81.8 °C

Note that moles of iron are used rather than grams. This is because of the units on the specific heat provided in the problem.


Problem #10: A 10.35 kg block of ice has a temperature of -22.3 °C. The block absorbs 4.696 x 106 J of heat. What is the final temperature of the liquid water?

Solution:

1) Raise the temperature of the ice to 0 °C:

q = (10350 g) (22.3 °C) (2.06 J/g¯1 °C¯1)

q = 475458.3 J

2) Melt the ice at 0 °C:

q = (6020 J/mol) (10350 g / 18.015 g/mol)

q = 3458617.818 J

3) How many Joules remain?

total used to this point = 3.934 x 106 J

total remaining: 4.696 x 106 J minus 3.934 x 106 J = 7.62 x 105 J

4) Determine temperature increase of liquid water:

7.62 x 105 J = (10350 g) (x) (4.184 J/g¯1 °C¯1)

x = 1.76 °C


Problem #11: How many grams of ice at -5.6 °C can be completely converted to liquid at 10.1 °C, if the available heat for this process is 4.74 x 103 kJ?

Solution:

1) Raise the temperature of the ice to zero Celsius:

q = (x) (5.6 °C) (2.06 J/g¯1 °C¯1)

2) Melt the ice at zero Celsius:

q = (6020 J/mol) (x / 18.015 g/mol)

3) Raise the temperature of the ice to 10.1 Celsius:

q = (x) (10.1 °C) (4.184 J/g¯1 °C¯1)

4) 4.74 x 106 J are used in total:

4.74 x 106 J = [(x) (5.6 °C) (2.06 J/g¯1 °C¯1)] + [(6020 J/mol) (x / 18.015 g/mol)] + [(x) (10.1 °C) (4.184 J/g¯1 °C¯1)]

The solution for x is left to the reader.

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

Some examples of problems covering four or five areas of the time temperature graph.

Go to the Time-Temperature Graph file

Return to Thermochemistry Menu