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Thermochemistry Example Problems

Example Number One
This problem is the *****alent of area number 3 on the time-temperature graph.

calculate the heat necessary to raise 27.0 g of water from 10.0 °C to 90.0 °C

The important factor about this problem is that a temperature change is involved. Therefore, the equation to use is:
q = (mass) (Dt) (Cp)
This summarizes the information needed:

Dt = 80.0 °C
The mass = 27.0 g
Cp = 4.184 Joules per gram-degree Celsius

Only one calculation is needed and it is:
q = (27.0 g) (80.0 °C) (4.184 Joules per gram-degree Celsius)
Multiply it out and round off to the proper number of sig figs.

By the way, you might want to think about how I knew to use the specific heat value for liquid water!!

Example Number Two
This problem is the *****alent of area number 3 and number 4 on the time-temperature graph.

How many kJ are required to heat 45.0 g of H2O at 25.0 °C and then boil it all away?

We must do two calculations and then sum the answers.
Calculation Number One uses this equation:
q = (mass) (Dt) (Cp)
This summarizes the information needed:

Dt = 75.0 °C
The mass = 45.0 g
Cp = 4.184 Joules per gram-degree Celsius

We then have:
q = (45.0 g) (75.0 °C) (4.184 Joules per gram-degree Celsius)
This gives an answer of 14121 J
Calculation Number Two uses this equation:
q = (moles of water) (DHvap)
This summarizes the information needed:

DHvap = 40.7 kJ/mol
The mass = 45.0 g
The molar mass of H2O = 18.0 gram/mol

Substituting, we obtain:
q = (45.0 g / 18.0 g mol¯1) (40.7 kJ/mol)

This gives 101.75 kJ. Adding 14.121 kJ and rounding off gives 115.9 kJ
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Example Number Three: 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: 4 calculations are needed - melt, raise, boil, raise.) Good luck.


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Solution
1) Melt
q = (33.3 g / 18.0 g mol¯1) (6.02 kJ / mol)
2) raise in temperature as a liquid
q = (33.3 g) (100.0 °C) (4.184 J/g °C)
3) Boil
q = (33.3 g / 18.0 g mol¯1) (40.7 kJ / mol)
4) raise in temperature as a gas
q = (33.3 g) (50.0 °C) (2.02 J/g °C)
The ChemTeam gets 103.7 kJ as the final answer.