Calorimetry - Calculating absorbed and released heat

[edimeo25]

Homework Statement

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

Homework Equations

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

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?

 

[Borek]

Looks OK.

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

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[Blueshift5]

Firstly, your calculations and approach seem correct. The difference between Q and "delta"H is that Q represents the total amount of heat absorbed or released in a process, while "delta"H represents the change in enthalpy. Enthalpy is a thermodynamic quantity that takes into account not only the heat absorbed or released, but also changes in pressure and volume. In this case, the heat absorbed by the water and the heat released by the burning ethanol are equal in magnitude but opposite in sign, resulting in a net change of zero in enthalpy. The molar heat of combustion, which you calculated to be 81.08 kJ/mol, is a measure of how much heat is released when one mole of ethanol is burned. Your result seems reasonable and in line with known values for the molar heat of combustion of ethanol. However, it is always a good idea to double check your calculations and units to ensure accuracy. Overall, your response shows a good understanding of the principles of calorimetry and thermodynamics.