[daisy007]

We did an experiment where we used energy from burning paraffin wax  to heat a known volume of water. We used the equation q = - cp(water) x m(water) x ∆T (water)
to calculate the amount of heat transfered to water. The mass of fuel burned was then taken into account to calculate the heat per gram of fuel consumed in the combustion. (∆Hc = Q/m)

The expected value is -41.5 kJ/g, but the result I got is much higher, even though it would be logical for it to be less than the expected value because of heat losses.

I was wondering if this is because I didn't take into consideration the can in which the water was heated?
How could I incorporate the can in the equation q = - cp x m x ∆T to get a more accurate result?

Thank you for helping

[thetada]

Hi Daisy,

Welcome to the forum. Can you show us the calculation you did when you were just including the water?

[daisy007]

These are the data that I used:

m  (can) = 14. 22 g
m  (water + can) = 162.58 g
m1  (C25H52) = 10.1270 g
m2 (C25H52) = 9.93 g
T1 (H2O) = 19.8ºC   
T2 (H2O) = 40.3⁰C

And these are the calculations:

Q = - cp (water) x m(water) x ∆T(water) = - 4.18 J/gK x 148.36g x 20.5 K = -12. 71 kJ

∆Hc = Q/m (paraffin wax) = -12.17 kJ / 0.197 g = -64. 53 kJ/g

[thetada]

Your calculations check out. I haven't done a calculation of energy transferred in which two different materials were involved. If you know what the metal was, you could look up the heat capacity and do a separate calculation alongside yours. You'd have to assume that the temperature of the can matched the temp of the water. But like your concern about lost energy, I think factoring in the can would only increase your value for energy transferred. You've calculated the energy needed to raise the temp of your mass of water. If you include the heat increase of the can as well, it's going to make the heat of combustion even higher. Unfortunately, I can't think of anything else to suggest. How certain are you that the fuel was paraffin?

[daisy007]

Thank you:) I'm positive the fuel was paraffin. I will try to do another calculation for the can.

[thetada]

Cool, tell us how the other calculation works out :-)

[Enthalpy]

The can adds its own heat capacity. Its temeprature is supposedly close to the water's one, so you add the mass*SpecHeat for the water and for the can.

And because the heat amount is then bigger, you'll compute an even greater combustion heat from your paraffin.

Something is wrong with your measures. Paraffins are near to heavy oil distillates like Diesel oil, which all offer 44MJ/kg combustion heat, not 65.