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Calorimetry - Calculating absorbed and released heat

  • Thread starter edimeo25
  • Start date
1. The problem statement, all variables and given/known data
Calculate the heat absorbed by the water in the calorimeter and the heat released by 1,0g of burning ethanol. Then, calculate the molar heat of combustion (kJ/mol) of the ethanol.

The calorimeter used was very makeshift: ethanol in an alcohol burner was lit under a can containing water, both of which were placed inside a hollowed out larger can.

Here is what is known:
mass of ethanol burned = 20g
mass of heated water = 280g
change in temperature = 30"degrees"C

2. Relevant equations

Q(water) = -Q(ethanol)
mc"delta"T = - (mc"delta"T)
280g*4.184J/g"degree"C*30 = -Q(ethanol)

3. The attempt at a solution
Q(water) = mc"delta"T = 280g*4.184J/g"degree"C*30"degrees"C = 35145.6J = 35.1kJ

Q(ethanol) = -Q(water) = -35.1kJ (for 20g)

Q(per gram of ethanol) = -35.1kJ / 20g = -1.755kJ/g = -1.76kJ/g

Q or "delta"H? (per mol of ethanol) = -1.76kJ/g * 46.07g/mol = 81.08kJ/mol

--> Does this make sense? Is it right? And what is the difference between Q and "delta"H?


Looks OK.

Check enthalpy definition. In some circumstances it is identical to Q, in some it is not.


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