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Q = c delta t......
Is this equation used to find the specific heat capactiy of the apparatus (where Q is the heat evolved in the reaction, and delta t is the temperature change in the reaction).
Carlo08
Well you wouldnt be expected to work out the specific heat capacity of the apparatus, thats if they actually have one :s
The apparatus will of course have a specific heat capacity, all materials do. However I don't see why you would want to find it. If you were to need the specific heat capacity of the apparatus then the equation would be something more along the lines of:
where m is the mass of the material in question, note that the apparatus would have to be all made of the same material, which makes me think that the question is a little spurious...
Isn't it where is the change in temperature?
The equation I've used is E = mc(delta)t
Q = energy used/needed
m = mass of thing heated
C = specific heat capacity of thing heated
t = change in temp
So, if you are given the start temp, the final temp, the mass and the specific heat capacity, you can find the energy needed/used.
If you're given all but one of the others, you can work out the remaining one by rearranging the formula.
Q = energy used/needed
m = mass of thing heated
C = specific heat capacity of thing heated
t = change in temp
So, if you are given the start temp, the final temp, the mass and the specific heat capacity, you can find the energy needed/used.
If you're given all but one of the others, you can work out the remaining one by rearranging the formula.
To find the heat capacity of a vessel, you carry out a known thermodynamics reaction, say the neutralisation of strong acid and base, for which you know the theoretical energy change.
When your measured energy change is lower for the solutions, you can factor the missing energy into the apparatus as a calibration.
example
For 50cm3 1M HCL + 50cm3 1M NaOH you would expect 57 x 0.05 = 2.85 kJ of energy released. This should cause the 100g of solution to increase in temperature by ( E = mcdeltaT)
delta T = 2.85 /(4.2 x 0.1) = 6.8 ºC
However, if your experiment only registers a 5º increase in temperature then you can say that 1.8º (using E='worth' of energy has been lost). This corresponds to 1.8/6.8 x100 percent of the energy lost = 26.5% which is 26.5/100 x 2.85 = 0.75 kJ
So your apparatus itself 'lost' 0.75 kJ for an increase of 6.8ºC
That means that it willl lose approximately 0.75/6.8 = 0.11 kJ per ºC change.
This is the calibration heat capacity (c') of your apparatus which can now be used for other experiments.
Total energy = mcdeltaT (solutions) + c'deltaT (apparatus)
When your measured energy change is lower for the solutions, you can factor the missing energy into the apparatus as a calibration.
example
For 50cm3 1M HCL + 50cm3 1M NaOH you would expect 57 x 0.05 = 2.85 kJ of energy released. This should cause the 100g of solution to increase in temperature by ( E = mcdeltaT)
delta T = 2.85 /(4.2 x 0.1) = 6.8 ºC
However, if your experiment only registers a 5º increase in temperature then you can say that 1.8º (using E='worth' of energy has been lost). This corresponds to 1.8/6.8 x100 percent of the energy lost = 26.5% which is 26.5/100 x 2.85 = 0.75 kJ
So your apparatus itself 'lost' 0.75 kJ for an increase of 6.8ºC
That means that it willl lose approximately 0.75/6.8 = 0.11 kJ per ºC change.
This is the calibration heat capacity (c') of your apparatus which can now be used for other experiments.
Total energy = mcdeltaT (solutions) + c'deltaT (apparatus)
theta = delta T
there is no 'delta theta'
there is no 'delta theta'
charco
To find the heat capacity of a vessel, you carry out a known thermodynamics reaction, say the neutralisation of strong acid and base, for which you know the theoretical energy change.
When your measured energy change is lower for the solutions, you can factor the missing energy into the apparatus as a calibration.
example
For 50cm3 1M HCL + 50cm3 1M NaOH you would expect 57 x 0.05 = 2.85 kJ of energy released. This should cause the 100g of solution to increase in temperature by ( E = mcdeltaT)
delta T = 2.85 /(4.2 x 0.1) = 6.8 ºC
However, if your experiment only registers a 5º increase in temperature then you can say that 1.8º (using E='worth' of energy has been lost). This corresponds to 1.8/6.8 x100 percent of the energy lost = 26.5% which is 26.5/100 x 2.85 = 0.75 kJ
So your apparatus itself 'lost' 0.75 kJ for an increase of 6.8ºC
That means that it willl lose approximately 0.75/6.8 = 0.11 kJ per ºC change.
This is the calibration heat capacity (c') of your apparatus which can now be used for other experiments.
Total energy = mcdeltaT (solutions) + c'deltaT (apparatus)
When your measured energy change is lower for the solutions, you can factor the missing energy into the apparatus as a calibration.
example
For 50cm3 1M HCL + 50cm3 1M NaOH you would expect 57 x 0.05 = 2.85 kJ of energy released. This should cause the 100g of solution to increase in temperature by ( E = mcdeltaT)
delta T = 2.85 /(4.2 x 0.1) = 6.8 ºC
However, if your experiment only registers a 5º increase in temperature then you can say that 1.8º (using E='worth' of energy has been lost). This corresponds to 1.8/6.8 x100 percent of the energy lost = 26.5% which is 26.5/100 x 2.85 = 0.75 kJ
So your apparatus itself 'lost' 0.75 kJ for an increase of 6.8ºC
That means that it willl lose approximately 0.75/6.8 = 0.11 kJ per ºC change.
This is the calibration heat capacity (c') of your apparatus which can now be used for other experiments.
Total energy = mcdeltaT (solutions) + c'deltaT (apparatus)
THANK YOU!
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