**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 CElectrical 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.