Four Equations Needed

**Example #1:** 33.30 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.)

**Solution**

1) Melt:

q = (33.30 g / 18.0 g mol¯^{1}) (6.02 kJ / mol)

2) Raise in temperature as a liquid:

q = (33.30 g) (100.0 °C) (4.184 J/g °C)

3) Boil:

q = (33.30 g / 18.0 g mol¯^{1}) (40.7 kJ / mol)

4) Raise in temperature as a gas:

q = (33.30 g) (50.0 °C) (2.02 J/g °C)

5) Add 'em up:

103.7 kJ becomes 104 kJ when rounded to three sig figs

**Example #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 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 kJto three sig figs, the answer is 319 kJ

**Example #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¯^{1}K¯^{1}

specific heat capacity for liquid water = 4.184 J g¯^{1}K¯^{1}

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¯^{1}K¯^{1}) = 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¯^{1}K¯^{1}) = 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 kJUse the rule for rounding with five to obtain 14.8 kJ for the final answer.

**Example #4:** How much heat is required to convert 87.8 g of solid ethanol at −114.14 °C to gaseous ethanol at 135.8 °C?

The molar heat of fusion of ethanol is 4.64 kJ/mol and its molar heat of vaporization is 38.56 kJ/mol. Ethanol has a normal melting point of −114.14 °C and a normal boiling point of 78.24 °C. The specific capacity of liquid ethanol is 2.44 J/g °C and that of gaseous ethanol is 1.43 J/g °C

**Solution:**

1) The first step is to describe what the ethanol will do as energy is absorbed:

1) melts at −114.14 °C <--- use molar heat of fusion

2) heats up from −114.14 °C to 78.24 °C <--- use specific heat of liquid ethanol

3) boils at 78.24 °C <--- use molar heat of vaporization

4) heats up from 78.24 °C to 135.8 °C <--- use specific heat of gaseous ethanol

2) Set up the four calculations outlined just above:

1) q = (87.8 g / 46.0684 g/mol) (4.64 kJ/mol) = 8.8432 kJ

2) q = (87.8 g) (192.38 °C) (2.44 J/g °C) = 41214 J

3) q = (87.8 g / 46.0684 g/mol) (38.56 kJ/mol) = 73.4900 kJ

4) q = (87.8 g) (57.56 °C) (1.43 J/g °C) = 7226.9 J

3) Add up the four enthalpies:

8.8432 kJ + 41.214 kJ + 73.4900 kJ + 7.2269 kJ = 130.7741 kJThree significant figures is the best choice, so 131 kJ is the answer.

Note how I changed the Joules in calculations 2 and 4 to kJ before adding.

**Example #5:** 40.0 g of ice at −16.0 °C is heated until all the water is converted to steam at 100. °C. How many kilojoules are required to do this?

**Solution:**

1) Identify what the water does as it absorbs energy:

1) heats up from −16 to 0

2) melts at 0

3) heats up from 0 to 100

4) boils at 100

2) Each behavior requires a calculation:

1) q_{1}= (40.0 g) (16 °C) (2.02 J g¯^{1}°C¯^{1})

2) q_{2}= (40.0 g / 18.0 g/mol) (6020 J/mol)

3) q_{3}= (40.0 g) (100 °C) (4.184 J g¯^{1}°C¯^{1})

4) q_{4}= (40.0 g / 18.0 g/mol) (40700 J/mol)Note that the enthalpy of fusion is in J/mol, as is the enthalpy of vaporization.

3) Work 'em out and add 'em up:

1) 1292.8 J

2) 13377.778 J

3) 16736 J

4) 90444.444 JTo three sig figs, the answer is 122 kJ

**Example #6:** How much heat is released when 95.0 g of steam at 100.0 °C is cooled to ice at −15.0 °C? The enthalpy of vaporization of water is 40.67 kJ mol¯^{1}, the enthalpy of fusion for water is 6.01 kJ mol¯^{1}, the molar heat capacity of liquid water is 75.4 J mol¯^{1} °C¯^{1}, and the molar heat capacity of ice is 36.4 J mol¯^{1} °C¯^{1}.

**Solution:**

1) Identify the behaviors involved:

(a) condenses at 100

(b) cools from 100 to 0

(c) solidifies at 0

(d) cools from 0 to −15

2) Since all of the constants use moles, we need to convert the mass of steam to moles:

moles = 95.0 g / 18.015 g/mol = 5.2734 mol

3) Each behavior requires a calculation (note the use of J rather than kJ):

(a) Condensing the steam at 100 °C:q = (5.2734 mol) (40670 J mol¯^{1}) = 214,469 J(b) Liquid water cooling off from 100 °C to 0 °C:

q = (5.2734 mol) (100.0 °C) (75.4 J mol¯^{1}°C¯^{1}) = 39,761 J(c) Freezing the water at 0 °C:

q = (5.2734 mol) (6010 J mol¯^{1}) = 31,693 J(d) Ice cooling from 0 °C to −15 °C:

q = (5.2734 mol) (15.0 °C) (36.4 J mol¯^{1}°C¯^{1}) = 2879 J

4) The answer to this problem is the sum of (a) through (d):

214,469 + 39,761 + 31,693 + 2879 = 288802 JTo three sig figs, the answer is 289 kJ