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![]() Thermochemistry Example Problems Example Number One This problem is the *****alent of area number 3 on the time-temperature graph. 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: 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. 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: 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: 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. ----------- . 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.
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دكتور عاطف خليفة كيميائي 500 امتحان كيمياء |
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