Specific heat is typically quoted in units of J/g/K or J/g/degreeC (Joules per gram per Kelvin or Joules per gram per degree Celcius). For example, if a substance has a heat capacity of 2 J/g/degreeC that means it takes 2 Joules of energy to raise the temperature of 1 gram of the substance by 1 degree Celcius. Note that when we are talking about temperature changes rather than absolute temperatures, a Kelvin is the same as a degree Celcius.

Why would you want to know the specific heat of a fluid? Well, suppose you want to Build your own flat panel solar thermal collector and you'd like to find or create a fluid that can store lots of energy for a small change in temperature. You might want to experiment with different fluid compositions to see what has the highest specific heat.

**Signing Up**

## Step 1: Equipment

- Digital postal scale
- Plastic cup that will hold at least 250ml
- Variable power supply]
- Digital thermometer with probe
- 7.5 ohm, 5W resistor] (or something close)
- Short length of wire
- Clock showing time in seconds (not shown)
- 250 ml of cold water (tap water will do, distilled is better) (not shown)

## Step 2: Setup

2. Put the cup on the digital scale and zero the scale.

3. Pour cold water into the cup until the scale reads at least 250 g. Record the mass "M" of water that you actually added.

3. Strip a short length of the insulation from the wire and connect the resistor in series with the variable power supply by twisting the bare wires around the resistor leads.

4. Put the resistor into the cup of water so that the resistor is submerged

5. Record the ambient temperature "Ta" and then put the temperature probe into the water as well.

## Step 3: Procedure

2. Record the water temperature every 2 minutes until the temperature is about 5 degrees C above the ambient room temperature.

Aside: Ohm's law states that Voltage (V), Current (I), and Resistance (R) are related by the formula V = I*R (where V is in volts, I is in Amps, and R is in Ohms). You measured V and I. Therefore you can calculate R = V/I and compare this with the known value of your resistor. In my case, V=7.5V, and I=1.00A. Therefore R=V/I = 7.5 Ohms which is the value of the resistor that I used. Hurray!

## 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 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.

## Step 5: Conclusion

The heat capacity of water was determined to be: 4.4 +/- 0.2 J/g/degree C.

This agrees with the known value of 4.2 J/g/degreeC

The same experiment could be used to measure the heat capacity of any substance that is a liquid at room temperature. For liquids that are electrically conductive, the resistor and wires should be insulated so that they do not come into direct contact with the liquid. Pure water is actually a poor conductor of electricity. It only conducts well when impurities (such as salt) are present. Some other household substances you might consider testing are: vegetable oil, olive oil, motor oil, rubbing alcohol.

**Sources of error**

Sources of error in this experiment include measurement error, heat transfer to the surrounding air, and cooling due to evaporation. Measurement error is likely to be the most significant of the three. With basic equipment, the specific heat can only be measured to an accuracy of about +/-5%.

**Improvements**

Accuracy could be improved by several methods:

1. Reduce error due to heat transfer and evaporation by using an insulated container such as a styrofoam cup with lid. I used what I had.

2. Reduce measurement error with better instrumentation. The voltage and current could be measured more accurately with a digital multimeter rather than relying on the power supply display. A higher resolution thermometer could also be used.

3. Reduce measurement error by increasing the voltage. Increasing the value of V will results in increased values of I and (T2-T1) as well. Since the absolute error in each of these quantities is fixed, increasing the values will decrease the percentage error. However, there is a trade-off since at higher or lower temperatures, the error due to heat transfer with the surroundings will increase.

4. Reduce measurement error and heat transfer error by curve fitting. Rather than simply using a datapoint 10 minutes before ambient temperature and another datapoint 10 minutes after, it would be more accurate to fit a polynomial curve to all the data points and calculate the rate of temperature change (the slope of the curve) at the instant the fluid reaches the ambient temperature. At this instant, there should be no heat transfer to or from the surroundings, so the rate of temperature change would be solely due to the input of electrical power.

**Further Reading**

For a detailed description of specific heat including known values for a variety of materials see Wikipedia's entry on Specific Heat Capacity.

For details of other interesting experiments and projects see my website IWillTry.org.