Electrical Energy.

Labs written for the CARA Space Explorers, Fall 1993

*
This is meant to be handed out to the students.*

Suppose you have managed somehow to separate a bunch of positive and negative charges. Since they attract each other they will want to flow back together. This flow is called an electric current and it is measured in amperes, or amps (A) for short. You can make this current do work, that is, you can extract energy from it, much the same way that you can make a flowing river do work to turn a water wheel, or in a hydroelectric generating plant. (Such a plant would turn the energy contained in the flowing water into electrical energy.)

We use electrical energy all the time. Most household appliances are powered by electricity. We often talk about electrical power. How is this related to electrical energy?

Power is the rate of using energy; it therefore has units of energy/time, or erg/sec. One Watt is equal to 10 million (10,000,000 = 107) erg/sec. The amount of energy would then be

electrical energy = E.E. = power x time E.E. = (erg/sec) x sec = ergFor example, suppose you have a refrigerator that uses 500 Watts of electrical power, and you operate it for one hour. The amount of energy that you consume is

(500 Watts) (10 erg/sec/Watt) x (1 hour) (3600 seconds/hour) = 1.8 x 10 erg.

This is the commodity for which Commonwealth Edison sends you a bill at the end of the month (except they use a unit called the "kilowatt-hour" instead of ergs).

It's sometimes easy to think of all the electrons flowing around in a circuit just like a stream. Just as a stream finds it easy to flow downhill, the electrons find it easy to flow towards the protons. Just like gravitational potential energy which measures how badly the stream wants to flow downhill, we can measure electrical potential to represent how badly the electrons want to join with the protons. We use a V to symbolize this potential. And, just like a stream might have some difficulty flowing over rocks on the way downhill, there is resistance against the electrons flowing down towards the protons. We symbolize this by an R. A power supply carries the electrons back up the hill, so to speak, so that the stream of electrons can run downhill again.

The power consumed by the electrons running downhill and getting carried back up is given by

E.P. = V/R

where V is the potential difference between the "top" and the "bottom" of the hill and R is the resistance to the flow of electrons. We will figure out how much power our resistor is using by measuring V and R.

We can determine the amount of heat produced by measuring the amount by which a certain mass of water heats up. If you recall, the thermal energy of a mass m of water is

T.E. = m C (T - T) ,

where C is the heat capacity (for water this is equal to 1 calorie/° C gram), T is the ending temperature, and T is the beginning temperature. Another thing you need to know is that 1 calorie is equal to 4.2 Joules or 42,000,000 (4 x 10) ergs.

E.E. = (E.P.) t,

where

E.P = V/R

is in fact equal to the amount of thermal energy that is produced by heating it. The resistor will be used to heat a mass of water, and we can measure the thermal energy that is deposited in the water by

T.E. = m C (T - T) .

Measure about 150 milliliters or 150 grams of water into a thermos.

mass of water m: (grams).

Measure the resistance of the resistor.

resistance of resistor R: __________________ ohms.

Place the resistor in the water so that it is completely covered, and arrange the thermocouple so that it is reading the temperature of the water (but not in direct contact with the resistor).

beginning temperature T1: (° C).

We will make a table of values for the temperature of the water for each interval of time that there is current in the resistor. The procedure is to turn the current on for one minute, turn it off, stir the water to spread the heat around, wait for 30 seconds, stir the water to spread the heat around, and then read the temperature. Then, repeat this one minute heating cycle ten times.

time (min) temperature ( deg C) temp. diff. T2-T1 (deg C) 0 1 2 3 4 5 6 7 8 9 10The resistor consumes a certain amount of power, or number of Watts. We can figure out how much by measuring the electrical potential across the resistor and also measuring the resistance of the resistor. Using the formula

P = V/R,

what is the power consumption of your resistor?

electrical power: (Watts)

(ergs/sec).

You now have enough information to compute, at each time step in you experiment, both the electrical energy and the thermal energy.

Before we do this, let's look more closely at the data to see if the water heats up uniformly with time. That is, we want to make a graph that shows, on the "x-axis," the time (0, 1, 2 , etc. minutes) and on the "y-axis" the difference in temperature compared to the starting temperature, T - T.

Make the plot using your data and a piece of graph paper.

What is the total electrical energy dissipated in the resistor?

E.E. = (erg).

To compute the thermal energy, use the formula

T.E. = m C (T - T).

where you have measured the mass m earlier, C was also given earlier, and "(T - T)" is the total change in temperature.

T.E. = (erg).

Compare the values you obtained for E.E. and for T.E. Are they close to each other? If they differ, can you explain why they might be different?

_____________________________________________________________

2. An apartment dweller is charged for 150 kilowatt-hours worth of electrical energy by Commonwealth Edison. The energy costs 10 cents per kilowatt-hour. How much is the bill?

3. Give an example of converting electrical energy into another form of energy (other than thermal energy).

4. A resistor is dropped in a thermos of water. The resistor is 1 ohm, and a voltage of 5 V is applied for 10 minutes. How much energy is dissipated in the resistor (Units!)?

5. A student heats up a mass of water with a submerged resistor. In the first experiment, the water heats up by 10 degrees and it takes 5 minutes to do so. In the second experiment, the same mass of water heats by again 10 degrees, but this time it takes 2.5 minutes. What is the difference in the electrical energy required in these two experiments?

**Important Disclaimers and
Caveats**

Go back to the Chicago Fall 1993 Energy home page.