ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: HarveyD on August 13, 2011, 11:22:43 pm
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Hey guys
We performed an experiment whereby we calibrated a calorimeter (by heating up water)
Just wondering what kind of questions would be asked on the SAC cause theres not much in textbook...
Anyone know any errors as well?
I was thinking water being splashed out might be one...
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Hahah I shall continue answering your questions :P
Basically, know how to determine the calibration factor, be able to apply this to measure different things, perhaps determine thermochemical equations.
As for errors, if it isn't properly insulated, too much/not enough stirring, not accurate volumes/measurements, faulty voltmeters etc.
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hmm yeah, i guess it would have to be more general question for this SAC rather than talking about observations
I'll ask my teacher just to make sure though
thanks for that :D
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with the insulation as an error, what effect would it have on the results?
Since change in T will be smaller, will it therefore decrease calibration factor and increase enthalpy value? (calculated values i mean)
also, if we were to sketch a graph on it, would the insulated calorimeter just remain constant after the temp rise, while non insulted one start decreasing (exponentially im guessing?)
and lastly, any other ways to minimise the source of error besides making sure the calorimeter is insulated? :P
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The calibration factor will increase if there is poor insulation, as it will take more energy to heat the inside of the container. However, you will use the calibration factor to work out your answer, but you can't tell how much heat you lose because you can't control it! It does produce a random error because you cant measure what's lost. Graphically its the same, because you can't draw that precisely either.
Another possible source of error is if you don't stir it. The heat won't evenly distribute itself throughout the container.
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hmm but when we did the prac, we kept the energy constant for all 3 trials (using same voltage)? so it wouldn't have any effect?
wouldnt an insulated calorimeter reach a higher temp than a non-insulated because energy will continuously be lost in the non insulted?
:S
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can anyone confirm? :/
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hmm but when we did the prac, we kept the energy constant for all 3 trials (using same voltage)? so it wouldn't have any effect?
wouldnt an insulated calorimeter reach a higher temp than a non-insulated because energy will continuously be lost in the non insulted?
:S
Remember that when calculating the Calibration Factor, it is Joules divided by change in temp. Hence;
Poorly insulated = Not much temp increase = Higher CB (lower denominator)
Well insulated = More temp increase - Lower CB (higher denominator)
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ahh yeah, my bad
so would an insulated calorimeter increase the enthalpy change (since change in t is larger and t is the numerator for the delta H formula)?
Also would i be correct in saying that if I sketched a temp vs time graph for an insulated and poor insulted calorimeter, the well insulated one would remain constant after heating is finished, while the poorly insulated would increase then drop away after heating has stopped?
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hmm so how would the enthalpy change if it wasnt well insulated?
Wouldn't the affected calibration factor be negated by the affected temperature change? (since they're both on the top in the formula for delta H)
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Change in enthalpy differences will be the same as the calibration factor, as it is CB per mol. Since the mol calculation will be the same for both, it'll be the same..
Poorly insulated = Not much temp increase = Higher CB (lower denominator) = Higher Enthalpy
Well insulated = More temp increase - Lower CB (higher denominator) = Lower Enthalpy
Woops, my bad, forgot the enthalpy EQ :-\
I can't think off the top of my head, you'd probably have to do a couple of calculations to see if there is a predetermined effect, however, the change in temp is the denominator on the CB, but on the top for enthalpy, so if my maths is right the effect would cancel out?
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ohh okay so when we are using the enthalpy change formula for the second reaction, the effect of the insulation on the temperature change can be considered negligible? ( delta H = CF x change in temp / number of mol )
Only the CF determined will affect our calculations
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See edited post above.
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dont think it wouldnt cancel out, cause the enthalpy change formula is used for the second reaction which would have a different change in temperature to the initial one which we used to calibrate it
its kinda confusing lol, hopefully my teacher only asks us about errors relating to the calibration factor
probably just say parallax errors when reading off the thermometer if she doesnt
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dont think it wouldnt cancel out, cause the enthalpy change formula is used for the second reaction which would have a different change in temperature to the initial one which we used to calibrate it
its kinda confusing lol, hopefully my teacher only asks us about errors relating to the calibration factor
probably just say parallax errors when reading off the thermometer if she doesnt
Ha my bad again, I'm just gonna stop talking now, I need to go to sleep!
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haha alright :P
btw does anyone know if adding too much/more reactants will affect the enthalpy/calibration factor?
it'd just be in excess so it wouldnt have an effect right?
just asking cause i was reading up on the prac and i came upon this info
Take care to note:
the effect on calculated calibration factor if too much of one of the reactants has been added
the effect on calculated enthalpy value of a reaction if one reactant is added more.
but I dont really know why...
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Adding more of both reactants won't do anything because essentially your trying to find the enthalpy value which you would only use ratios to get the correct value for. E.g. 1 mole produces 10kj then 2 mole produces 20kj
If your only adding excess of 1 reactant then of course use the limiting reactant.
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ah k thats what i thought
for the reaction of NaOH and HCl
if you were to decrease the number of mol equally (i.e. decrease the volume for both)
would the heat of neutralization change?
i was thinking that it would increase from the formula, but then wouldnt decreasing the volume also decrease the temperature rise proportionally, meaning that it wouldnt change at all :S
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ah sorry to bother everyone again :S
but would anyone be able to check my explanations for errors relating to this experiment
CF (Too High)/ Would also apply to enthalpy being less then expected -
1) Poor Insulation – If the calorimeter is not properly insulted this, it will result in a loss of heat to the surroundings, hence it will reach a lower temperature than it should. This will therefore mean that the change in temperature will also be lower, and since this is on the denominator for the formula of the Calibration Factor, a poorly insulted calorimeter will thus lead to a higher calibration factor.
2) Parallax Errors – If the temperature readings are taken from above, then this will result in a lower temperature being recorded after the reaction has taken place. Hence this will mean that the change in temperature obtained will be lower and since this value is on the denominator for the formula of the Calibration Factor, such parallax errors will thus lead to a higher calibration factor.
3) Uneven stirring/slow stirring
CF (Too Low) / Enthalpy more then expected
1) Parallax Errors – Basically the same explanation for too high, except when readings are taken from below etc
Can’t think of another others for this :/
I guess that means there’s a less chance of it happening...
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Contamination, incomplete rxn etc
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Dodgy wiring in the electrics?