## Step 4: Results

Data
`   Mass of water, M        = 251 g   Ambient Temperature, Ta = 20.1 degrees C   Voltage, V              = 7.5V   Current, I              = 1.00A      See a plot of the data below:   t [seconds] - T [degrees C]   0           - 13.8   120         - 14.9   240         - 15.9   360         - 16.8   480         - 17.7   600         - 18.6   720         - 19.4   840         - 20.2 (close to ambient temperature)   960         - 20.9   1080        - 21.7   1200        - 22.6   1320        - 23.4   1440        - 24.0   1560        - 24.8`
Calculations
The ambient temperature was 20.1 degrees C. One of the datapoints was 20.2 degrees C which is very close to ambient. To minimize error due to heat transfer to or from the surroundings, lets look at the data from 10 minutes before till 10 minutes after this datapoint.
`   t1=240,  T1=15.9   t2=1440, T2=24.0   total time         = t2-t1 = 1200 s   temperature change = T2-T1 = 8.1 degrees C`
Electrical power in Watts is equal to V*I or voltage times current. Also, 1 Watt is equivalent to 1 J/s (Joule per second). Thus:
`   Power P (J/s)               = V*I   Energy E (J)                = P * (t2-t1) = V*I*(t2-t1)   Specific heat (J/g/degreeC) = E / M / (T2-T1) = V*I*(t2-t1) / M / (T2-T1)`
Substituting measured values gives:
`   Specific heat  = 7.5*1.00*1200/251/8.1= 4.4 J/g/degreeC`
Accuracy
The quantity V has a tolerance of +/-0.1V or about 1.3% (0.1/7.5).
The quantity I has a tolerance of +/-0.01A or about 1.0% (0.01/1.00).
The quantities t1 and t2 each have a tolerance of +/- 1 second.
Therefore the quantity (t2-t1) has a tolerance of +/- 2 seconds or about 0.2% (2/1200).
The quantity M has a tolerance of +/- 1g or about 0.4% (1/251).
The quantities T1 and T2 each have a tolerance of +/- 0.1 degrees C.
Therefore the quantity (T2-T1) has a tolerance of +/-0.2 or about 2.5% (0.2/8.1)

When multiplying or dividing quantities, their percentage tolerances simply add together. Therefore, the calculated value of specific heat has a tolerance of about 5.4% yielding a final result of 4.4 +/- 0.2 J/g/degreeC.
<p>How to measure the Cp of air??</p>
There is no such thing as a "degree Kelvin". Only Kelvin.
Thanks. Fixed. Did you created your user account just to tell me that?
Yes, all in the name of science!
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meal
what?
exactly...
what?
Just as I suspected...
...You are...umm...EATING!
OK...I'll tell you what it means...the suspense has been building long enough...OK, are y'all ready for this...OK...brace yourselves... OK, I first said it was lame, as you can see above, where it says "lame", but then I re-examined the instructable and found it to be not so lame, so I had to 'eat my words' so to speak, so if I was going to have to 'eat my words' as I have said, then I was going to 'make a "meal" of it', so to speak, and, conveniently, the letters of lame, rearanged, spell meal, so I did not even have to cook up a new message to make a "meal" of it, rather just use the ingredients I had on hand, mixing up the 'l', the 'a', the 'm' and the 'e' to produce a "meal" clever, no?
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im just glad aliens took over pluto
I learned this stuff in chemisty using heated metals. Great instructable coverage.
Nice! Very nicely done. Kudos to you on format and explanation. I like how you relate this instructable to the solar water heating instructable. I am also really impressed with your final step where you cover error sources, improvements and further reading. A+