If energy is like money in the bank, then energy conversion is like withdrawing funds to make a specific purchase. Previously, we used electrical energy to heat water, and noted the efficiency of this operation. In this lesson we will again use electrical energy to heat water, this time until we boil the water and convert it into steam. This takes more energy because we must change the state of the water molecules into a higher energy form - steam. For each gram of liquid water at its boiling point that we convert into steam at the same temperature, we need 540. calories of energy. This is called the latent heat of vaporization or latent heat of boiling for water. We will use a microwave oven to heat our water, and note the efficiency of this method compared to the immersion heater used to make tea in the previous lesson . Which do you think will use less energy to heat the same amount of water through the same temperature change? Why?
Figure 1:
Equipment for Energy Conversions II balance beaker electric burner graduated cylinder microwave oven multimeter pressure cooker stopwatch
Styrofoam ® cup thermometer
[At the end of this lesson students will be able to...]


 use latent heat of boiling in determining total energy input to boil a given mass of water. describe the steps needed to collect data and calculate the efficiency of heating water on an electrical hot plate. describe the steps needed to collect data and calculate the efficiency of heating water in a small microwave oven.
 extend the steps above to determine total energy expended in both heating liquid water and then boiling the water.
 work problems requiring energy conversion and application to heating water and boiling it.
1.
Suppose we make tea by heating water on an electrical hotplate. Do we use any less energy per quantity of water heated than we used with the immersion coil in the previous lesson ? Predict the results.
______________________________________________________________________________
Energy Conversions II -- Page 1 of 6

2.
Suppose we make tea in a microwave oven. Do we use any less energy per quantity of water heated than with the hotplate or immersion coil? Why do you think that?
3.
Is there any difference in the energy needed to heat 100. grams of liquid water from
20.°C to 100.°C and the energy needed to boil the same amount of water at 100.°C? Why to you think it is the same or different? Does this make any sense from a molecular viewpoint?
Use electricity to heat water into steam
We measure the electrical energy by monitoring the electrical current (ampere) over time.
The product of voltage, current and time yield energy (Volt × Amp × Sec = J and Volt × Amp =
Watt). Part of this electrical energy is converted into thermal energy by boiling water to form steam. The beaker will be carefully weighed to determine the amount of steam formed during the demonstration. The thermal energy equals the water loss times the latent heat of vaporization.
The efficiency of converting electrical energy into steam is quite high.
First conversion: Electrical to thermal
Voltage of heater =
Current of heater =
Power of heater =
Duration of boiling =
Electrical energy consumed =
Weight of water + pressure cooker before boiling =
Weight of water + pressure cooker after boiling =
Amount of steam produced =
Heat of vaporization of water =
Energy in vaporization =
Efficiency of boiling =
Note that for phase transitions there is no temperature change, and the equation relating energy
E , mass m , and latent heat of phase transition (solid/liquid or liquid/gas) L is:
E = m × L
Generally, for a given substance the latent heat of fusion (solid/liquid phase transition) differs from the latent heat of vaporization (liquid/gas phase transition), and the latent heat of fusion is often represented by L f
while latent heat of vaporization is represented by L v
.
______________________________________________________________________________
Energy Conversions II -- Page 2 of 6

The electromagnetic spectrum
Energy can be transmitted as photons of radiation, and the energy of individual photons is defined according to the electromagnetic spectrum. This spectrum ranges from the highestenergy forms of radiation (gamma rays, with the shortest wavelengths and highest frequencies) to the lowest-energy forms of radiation (radio waves,
Figure 2: The Electromagnetic Spectrum with the longest wavelengths and lowest frequencies). Visible light is simply a form of electromagnetic radiation, with moderate wavelengths and thus moderate energy. A microwave oven produces radiation in the microwave range, which is less energetic than visible light or infrared radiation but more energetic than radio waves.
Why does radiation in this band work so well for cooking food? The answer lies in the nature of water, the major component of our food (and us): water molecules are very efficient at absorbing radiation in the microwave range.
Let's determine the efficiency of a microwave oven heating by again monitoring the electrical energy consumption of the microwave and the heating of water. The conversions are from electrical energy into microwave energy and then to heat (mass of water times temperature rise times specific heat). Notice the unit of conversion from electrical energy units (J) to thermal energy units (calories).
Power to the microwave oven =
Duration of heating =
Electrical energy consumed =
Mass of water used =
Temperature rise =
Specific heat of water =
Energy absorbed by water =
Efficiency of the microwave oven =
______________________________________________________________________________
Energy Conversions II -- Page 3 of 6

Can we extend this experiment with the microwave oven to account for both heating of water and boiling?
Power to the microwave oven =
Duration of heating (need to heat longer to induce boiling) =
Electrical energy consumed =
Mass of water before heating =
Initial temperature of water =
Mass of water after boiling =
Specific heat of water =
Latent heat of boiling for water =
Energy absorbed by to heat to boiling point =
Energy used for boiling =
Total energy transferred to water for boiling =
Efficiency of the microwave oven =
1.
Did any of these results surprise you? Which ones? Why?
2.
Does it matter what type of container we use to hold the water when we compare efficiencies of heating? Explain your reasoning. Can this be used to explain any error in our results? Do you think Styrofoam ® and aluminum or stainless steel absorb the same amount of heat?
3.
If you are given 100 grams of an unknown liquid and told that you must heat it to boiling, show the steps that you would use to determine its latent heat of boiling. All critical data measurements must be shown in your method. At what point would you stop adding heat?
I recently called the Alabama Power Company in Auburn and talked with their consultant about electric hot water heaters. She said that for my family of 3 people using two bathrooms and one washing machine our household score of "6 points" indicates that we need 50. gallons of hot water capacity. The Rheem Model 81V52D 50-gal. electric hot water heater which Alabama
Power sells for $239.00 has a maximum power rating of 4500. watts for the two immersion heating elements it contains (2250. watts @ 120 volts each). She recommended that I set the thermostat for 120. degrees Fahrenheit so I could have nice hot showers and get my clothes clean. The average year-round water temperature of the cold water coming into the hot water heater is 60.0°F.
Assume that my family uses an average of 50. gallons of hot water per day, and that my hot water heater must heat all of this water from 60.0°F to 120.°F before it leaves the hot water tank.
______________________________________________________________________________
Energy Conversions II -- Page 4 of 6

Assume that both elements are drawing their full rated power until the cold water reaches the thermostat setting. Assume that my hot water heater is 90.% efficient in converting electricity to hot water, and that no heat is lost from the tank or pipes taking it to my shower or washing machine. Electricity in Auburn costs $0.075 per kilowatt-hour. One gallon of water weighs 3.782 kilograms. One Celcius degree temperature change equals 1.8 Fahrenheit degrees temperature change. One kilowatt-hour equals 853 kilocalories. a.
What is the expected daily cost of heating all of my home's hot water? b.
How much time it will take my water heater to "recover", i.e. get 50. gallons of water back up to 120.°F. after I take two showers and wash a load of clothes - all at the same
"instant"? c.
How many kilowatt-hours are used to heat my water each year? d.
What is the expected annual cost of the electricity to heat my home's water? e.
Does the first year's electricity cost exceed the price of the Rheem Model 81V52D heater? If Alabama Power is making a 50.% profit on sales of electricity, would it make sense for them to give away hot water heaters to replace gas versions?
______________________________________________________________________________
Energy Conversions II -- Page 5 of 6

Imagine that you are a romantic in France and you bought an air conditioner in Paris (it uses 220.
V). The output power of the compressor is rated at 1.0 horsepower and the efficiency of the compressor at 40.%. But you are a very scientifically-minded romantic and you decided to measure these values yourself. You find that the output horsepower is correct but the efficiency is only 20.%.
(a) Calculate the current draw of the air conditioner. Hint: first calculate the two horsepower values needed for the two efficiencies. Then calculate the current needed for those two horsepowers.
(b) Speculate on the danger you are in (remember that there are current ratings for different conductors).
______________________________________________________________________________
Energy Conversions II -- Page 6 of 6