Thermochemistry Problems:
Three Equations Needed

Problems using one part of the T-T graph

Problems using two parts of the T-T graph

Problems using four parts of the T-T graph

Problems using five parts of the T-T graph

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Problem #1: 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 #2: 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 #3: 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 #4: 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 #5: 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 #6: 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 #7: 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.


Problems using one part of the T-T graph

Problems using two parts of the T-T graph

Problems using four parts of the T-T graph

Problems using five parts of the T-T graph

Go to the Time-Temperature Graph file

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