Precautions
1. To properly place the lid, otherwise heat lose might increase, to <minimize heat loss as much possible using insulation.
2. To maintain uniform temperature make sure to keep stirring the calorimeter.
3. Make sure to keep constant energy supply by heater – 12V.
Variables
Independent – Is the power heating the up the water. This is because if you change the power, then the change in temperature will also change proportionally. So you have to keep it constant throughout, otherwise this might lead to un-reliable data.
To manipulate you could change the power, by increasing the current or the volts. So you could change the current from 12V, to 16V, which would change the power. But this has to be done at the start of the experiment, and then kept constant throughout.
Dependent – This is the change in temperature, as this is dependent on the power supplied to the heater, if changed then the change in temperature will also change. To manipulate, you will have to change the Independent variable.
Controlled – This is the mass of the water, the time intervals and the power. These have to be kept constant, or otherwise it won’t be a fair test. The mass of the water is kept at 0.425 kg, 120 second time intervals and 12W power supply. To manipulate you could increase the mass by 0.010 kg, or increase the power supply to 16W, or even increase or decrease the time intervals to either 300 second or 30 seconds. But the changes have to be kept constant otherwise.
Results
Calculations
To find out the Specific Heat capacity for water we will have use this formula: Q = M x C x θ T.
First we need to find out how much heat is received (Q).
Power = Energy Transferred / Time taken
E =Q / T
Where:
E = the power in watts
Q = the energy in joules
T = time in seconds
12 x 0.85 = Q / (30 x 60)
Q = 10.2 x 1800
Q = 18360 joules
Now to find the Specific Heat capacity we have to go back to the original formula, and deduce these finding into it. So;
Q = 0.425 x C x 10 Re-arrange the formula
C = Q / (M x Δ θ)
C =18360 / (0.425 x 10)
C = 4320 j/kg˚C
We can check this by using the graph to find the specific head capacity; again we will use the formula: Q = M x C x Δ θ.
P x T = M x C x Δ θ
P = M x C / T
P = M x C x Gradient
C = P / M x Gradient
C = 12 /(0.425 x 0.0066666667)
C = 4235 j/kg˚C
Conclusion
From my calculation and results I have obtained the Specific Heat Capacity of water to be 4320 j/kg˚C for this practical. Now I can argue that this is not a 100% accurate experiment as the scientifically defined Specific Heat Capacity of water is found to be 4200 j/kg˚C.
There is a difference of 1320 j/kg˚C, from my results and the initial value; but I used two different ways to calculate the answer. I used the formula, which gives 4320 j/kg˚C, but then I also used the gradient which in result gave me a more accurate answer of 4235 j/kg˚C.
As you can see the gradient gives a much more accurate answer, this is because when you use the method of calculating the sphc, by subtracting the final temperature by the initial temperature, you only base your calculation of two sets of data. This is unreliable, because if one of them proves to be inaccurate, then this impacts the answer proportionally. The gradient method is more accurate, because you are taking the data from the line of best fit, which allows you to predict the value of one variable based on a value of the other; avoiding erroneous sets of data.
Using the gradient method I got an answer of 4235j/kg˚C, which is very close to the initial value of Specific Heat Capacity of Water. There is a difference of only 35j/kg˚C, and a difference of 85j/kg˚C from the calculations using the other method.
So why isn’t my answer identical to the genuine calculation? Well this can be caused by many reasons, one is that the calorimeter partially absorbed some of the heat which in turn meant, that it needed more energy to raise the temperature of the water, as the particles had less energy due to heat loss. Another reason could be because it was insulated 100%, which meant that it lost some heat to the air, which again meant that the particles didn’t have as much energy so it needed more. Though this reasons will only slightly affect the answer, as we didn’t not perform the experiment at a very high temperature.
In my hypothesis I stated that my answer for the specific heat capacity will differ slightly from the original, and I was right, but I also stated that the change in temperature will be constant, and that the temperature will rise uniformly. Theoretically I should be right, but from my results you can see that the change in temperature wasn’t constant. This is because I didn’t stir, the calorimeter enough and persistently. This would have made the particles in the water move more so the temperature would have then increased. The more stirring that was done could have allowed the temperature to increase at a fully uniform level. And I think that the main reason why my results of the specific heat capacity doesn’t exactly equal 4200 j/kg˚ C.
Evaluation
I think that I was very close to the precise specific heat capacity of water. With a difference of only 35j/kg˚C I think that there are variables, which cannot be controlled 100%. But I think I could have improved this experiment. I think next time I could take a wider range of data, for the temperatures. This means to repeat the experiment numerous times. This will allow me to have more accurate data. Also I think I should keep the stirring of the water more persistent next time. I could either, have a plan of something like, stirring for 2 minutes, and then stopping for an minute etc. Or I could persistently keep on stirring through the whole experiment; this will allow the particles to move around the beaker more, making it a fairer test.
Also I could increase the power supplied to the calorimeter, which would allow me to analyze the relationship at higher temperatures, and also more modern equipment.
All these factors could be improved next time, hopefully giving me even accurate results.
