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Example Number One
This problem is the equivalent 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) (Δt) (Cp)
This summarizes the information needed:
Δt = 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 equivalent 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) (Δt) (Cp)
This summarizes the information needed:
Δt = 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) (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
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.
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. Go to answer.