Download Files
 Site Index

 Top of this Page
 More Information 

 Freeze Menu 

Practical Information

    What Is the REAL POWER of Your Microwave Oven?

    A Microwave oven's cooking power is rated in Watts, that is, the amount of microwave power it delivers to the food. This is not to be confused with the power the oven draws from the power outlet on the wall. Most ovens draw about 1200-1500 watts electrical power, and produce varying amounts of microwave power to cook the food. Some are very small, specifying cooking power of only 500 or 600 watts, while others are rated at over 1100 watts. Do you know what power your oven really is? Worse, as ovens age, the oven's magnetron cavitates less efficiently and the power drops off.

    You'll have two methods for measuring your oven. The simple method should be accurate within maybe 20%; the more exact method is just that, exact!

    And, these methods use information found elsewhere on this site! Consider the possibilities!

    Items needed:
    1 working microwave oven
    8 oz. Pyrex or Oven-proof glass measuring cup
    Faucet Water
    Thermometer, optional
          (at least 50-180° F or 10-80° C) for the exact method
    Watch or Stopwatch, optional
          (it's possible to use the timer on the microwave)

    Simple Method:
    Fill the cup with precisely 8 oz. of cold water from the faucet and place the cup near the center of the microwave oven. Use a stopwatch or carefully note the time, and start the microwave oven by setting it on High for about 5 minutes. Watch very closely for the water in the cup to boil! We do not want to heat the water beyond the point it begins boiling.

    Warning! Read about Superheated water!!

    This is the most difficult part, determining precisely when the water boils. First, a few bubbles will begin to form, then a few will rise to the top. When more begin regularly rising together, click the stopwatch and get the elapsed time in seconds, and stop the oven. You don't want to wait for a vigorous, rolling boil!

    Now use this simple formula: Power in watts = 83564 / secs
    (If your time was 70 seconds, then 83564 / 70 = 1194 Watts)
    There are two huge problems with the simple method, you are never sure exactly when the water begins to boil, so when it first reached 212° F. is not clear. Additionally, you didn't measure the water temperature from the faucet - in the winter it is typically below 50° F and in the summer above 60°. The simple formula assumes it was 60°. For more accuracy, you could add or subtract 550 from 83466 for each degree the water was above or below 60.

    But why? Just move on the exact formula below.

    Exact Method:
    The plan is the same, fill the cup with exactly 8 oz. of water and with this method, measure the initial temperature of the water. Heat the cup of water in the oven on High for at least a minute. You want to raise the water temperature to at least 150-180°, but NOT TO THE BOILING POINT! You can use the 'Expected Boiling Times' chart below to estimate the heating time you should initially use to set the oven.

    Warning!! Read about Superheated water!!

    If you have a powerful oven, it will reach that temperature range in about a minute, a very small oven will take almost three minutes. Start with a shorter guess and work up.

    After your heating time period is up, immediately open the oven and measure the temperature of the water. Now you have the initial water temperature in degrees F, the heated temperature, and the time spent heating.

    If you need to convert the temperature from Celsius to Fahrenheit (Use your 'Back' button to return here afterwards.)

    Now use this exact formula: Power = temp rise X 549.12 / seconds
    Temp Rise is the final water temperature minus the starting temperature. (Ex. 50° F starting temperature, 190° heated temperature, and 77 seconds heating time. (190-50 = 140° temp rise) So 140 X 549.12 / 77 = 998 Watts.)

    Now you know your microwave oven's power. You should check it periodically to determine whether your oven's power is dropping, and how fast. And it will help you properly adjust your cooking times.

    Most cooking instructions give a range of times to microwave their product. If only a single time is given, it may be aimed at 700 Watt ovens - the original industry standard. If your oven produces 1000 Watts or above you should generally use the shortest suggested microwave cooking times. 700-800 Watts needs the longest suggested times. Less than 700 Watts and you'll need to guess about even longer times!!

    Some Expected Boiling Times, assuming 8 oz. of 60° F water:
    70 sec = 1200 watt
    83 sec = 1000 watt
    104 sec = 800 watt
    139 sec = 600 watt
    167 sec = 500 watt
    180 sec = 464 watt

    So, would you like to know how easy these formulae were to develop?
    One could assume it would be very difficult, because elsewhere on the web, the best help most other sites can offer for testing a microwave oven's power is to suggest heating 8 oz. of water for 3 minutes, and if it boils the water, the oven must be working! Notice the infamous '3 minute' boiling test time simply means the oven can produce at least 464 watts. Real Good Info.

    Merely heating the food indicates the microwave is somewhat working. Why, I've got an old microwave out back that's not plugged in, but it still defrosts frozen food....... eventually. For food safety, you need to know how well your oven is working!

    With the information found on this site, it's quite easy to develop an accurate test. From the Conversion Information page, 1 Gal = 128 oz. From the Energy Information page, 8.34 BTU raises 1 Gal. of water 1° F. and 1 KilowattHour = 3413 BTUs. So, we know how many ounces are in a gallon of water, how many BTUs raise a gallon of water a degree and how many kilowatt/hours are needed to produce those BTUs.

    The basic formula is:
    Power (watts) = ((gallons X temp rise X BTUs needed to raise 1 gal. 1° F. X seconds in one hour / seconds used) / BTUs needed to produce one watt).
    Or: Power = ((gal X temp rise X 8.34 X 3600 / seconds) / 3.413)
    Or: Power = (gal. X temp rise X 8797) / seconds
    Or: Power = (oz. X temp rise X 68.72) / seconds

    We'll stick with that last formula simplification and re-figure our first simple example. We assumed the faucet water temperature was 60°, making a temperature rise of (212-60 = 152°). With 8 oz., (8 X 152 X 68.72 = 83564) / heating seconds of 70 = 1194 watts. Now you can see the origin of the simple formula.

    Warnings and Caveats
    This doesn't allow for the heating of the container, which should be insignificant if doesn't absorb microwave power itself, like high-lead pottery. The container will absorb some heat energy from the water, just as the surrounding air does, but this occurs while cooking as well, so consider it a constant factor. After all, we weren't trying to determine the time to boil water in a zero-G environment with no container. We want the results to apply to real-life cooking.

    Check you household Voltage Level to the oven. Ovens don't contain electrical regulators to keep the power constant with varying incoming voltage levels. Your line voltage at the wall outlet may vary from below 100 volts to above 120 volts

    Fact: When the voltage rises from 100 to 110 volts, you have 20% more power! From 100 to 120 volts, you have 40% more power!! Typically appliances are rated at 110, 115, or 117 volts, and most often 115-117 volts. So you would not often exceed the manufacturer's power value by having voltage appreciably higher than 117, but when the voltage fell to 100 or so, you would certainly have less power - about 20% less power. A 1000 watt oven operating with only 100 volts can produce only about 800 watts, and for all practical purposes, it is an 800 watt oven. And that's exactly what this is all about - How much power does YOUR microwave oven produce?

    Is there anything you can easily do to change the voltage? Usually, No!
    Sometimes the voltage varies during the day as the power company changes it. The other factors within your house tend to more consistent. But don't forget a summer afternoon, using the microwave while the air conditioner is running and your electric oven in the kitchen baking something may easily drop your voltage level. Those units running at the same time may drop your voltage, but it's usually around 5%. It's just another factor to consider if you are cooking something delicate in the microwave and need exact times. Of course, an unusually low voltage may also indicate a wiring problem.

    Microwave Oven Trivia

    Don't set your cooking time on the microwave for seconds not divisible by 10. That is, use 10, 20, 30, etc., but NOT 5 or 15. The reason? Your microwave turntable turns one revolution every 10 seconds and if you try to warm your coffee for 15 seconds, it will stop on the BACK of the turntable, rather than just inside the door, where you originally placed it.

    Don't use the exact center of the oven to place your things to be heated, they will be more evenly heated if they rotate around on the turntable, rather than sitting in just one spot.

    Do you know why your microwave oven can't brown food? It's because the microwaves affect the water molecules in the food by agitating them to the point where heat is produced. BUT the highest temperature it can raise these water molecules is to 212° F, then the water turns to steam and is released from the food. So, that becomes the limiting temperature and you can't brown at 212, you need 325-400° F.

    You may argue about items you improperly heated and perhaps they caught fire. Isn't THAT hotter than 212°. Yes, it was, but the items reached that temperature after all the moisture was removed from them. Do you REALLY want to eat any food with NO moisture in it? I didn't think so.

    Yes, and you may further argue about the "browning trays" used in microwave ovens to brown food. OK, you got me. Something came out of the oven, browned. Although it was the tray that browned it, after the microwave heated the tray. And because the tray had no moisture to vaporize, the microwave heated the tray to a higher temperature. So, if you heat a cup of water to boiling and pour it on your flowers, was it the microwave that killed your flowers, or did it just heat the water that killed them?

    Superheated Water

    Under normal conditions, you can only heat water to 212° F., then it begins boiling. Once boiling, any more energy delivered to the water only makes it boil more vigorously. Superheating is an unusual event that can happen to anyone at anytime, although it never seems to happen to bad people, only to nice people like you and I.

    Normally a pot of water on the surface of a stove begins to boil as the temperature of the water at the bottom reaches 212° F. Then as steam bubbles begin to form, they rise to the top and are released as steam. A microwave oven heats all areas of the cup of water somewhat evenly. It is possible to heat a liquid hotter than 212° F at normal atmospheric pressure, and the liquid NOT boil! Of course, if you continue heating it, it usually begins to eventually boil. But when it's above the boiling temperature, the water is called 'superheated'. And just the slightest disturbance of the surface causes the water to suddenly and violently turn to steam. When an amount of water is turned to steam, the vapor expands roughly 1000 times the volume of the water! It's not unlike the effect that drives geysers. Part of the magic allowing the superheated condition is a container with a very smooth surface (like glass) that doesn't hold small or microscopic air bubbles when filled. If any bubbles are formed, and released, when they break the surface, the boiling begins and prevents superheating.

    If a person reaches into the microwave oven and picks up the cup with the superheated liquid, they may or may not be able to clear the microwave oven before they are severely burned with second and third-degree burns from the water. This is why most recent microwave instructions include '...then allow it to sit for 2 minutes before removing'

    Remember the effect of the line voltage on the microwave's power? If someone normally needs 1 minute to heat their coffee just below boiling, then a 10% higher line voltage from the wall socket can push the coffee into the superheated region, making it a trap just waiting for you to touch it.

    Another way you can accidentally make superheated liquids is to start heating a cup of coffee, get interrupted with perhaps a phone call, eventually return and reheat the coffee. If you waited longer than 30-40 minutes, the almost boiling coffee you nuked in the first session may have cooled sufficiently. Any less time, and you will be guilty of 'stacking' - you reheat for your 'normal' time, but with coffee already semi-hot - you really only needed a few seconds to heat it back to near boiling, yet using your full burn time superheats it. Surprise!

    How can you protect yourself? After heating a liquid, either allow it to cool for at least 30 seconds to see if automatically progresses from a superheated condition to boiling, or whether the temperature drops below the potential danger zone. OR, if you are in a hurry and can't wait - disturb the surface, either by touching it with a spoon or bumping the cup with a spoon or pencil to see if it suddenly explodes. Maybe you can get your arm back out of the oven in time to prevent getting scalded. Maybe not. Your call.

    Most discussions of superheating all assume it only occurs in microwave ovens and labs, never with a standard kitchen stove. Wrong! - Been there, done that!

    I once assumed water, at one atmosphere pressure, would always boil at 212° Fahrenheit. So, to verify the temperature settings on a standard kitchen oven, I put a small Pyrex bowl filled with water in the oven and set the temperature for a very low value - just below 200°. I let the oven come up to temperature, checking the water status, while peering through the glass oven window using the oven's internal light. Then I began raising the oven temperature in small increments - when I got above 212, I assumed I would see the water boil - just like casseroles bubble and boil!

    Time passed. MUCH time passed as I slowly raised the dial. I felt some satisfaction knowing my original suspicion was correct, the damn oven dial was incorrect. I started using larger increments, deciding I would 'bracket it' and zero in on a second attempt. But first I needed to get some boiling started.

    As I reached the 500° mark, I knew something was amiss. The water was not boiling, yet the oven was bristling with heat, hot enough to feel several feet away from the oven. At this point I actually opened the oven door to look directly at the water, just in the event the image of the cool, unboiling water had somehow etched itself into the glass of the oven door, preventing me from seeing the real water. No such luck, the water was still there, and not boiling. I turned the oven off and left the bowl in place (thank goodness!!!!!) because it was obviously very hot. This was just another failed experiment of mine - another 'learning left in the lurch' example. But the search for the reason for that failure haunted me for a while afterwards.

    Finally I latched onto the correct reason, almost. I theorized, as many microwave oven superheat enthusiasts do, the water didn't boil because it was evenly heated from top to bottom. On the stove top, the heat is transferred to the bottom of the pot (and some to the sides), and there is always some gradient of temperature from bottom to top. As the bottom is heated to 212F, the water rises through the colder water and is slightly cooled while the colder upper water replaces it. I also assumed the temperature of the rising water would stay at 212F until it reached the surface, where it would vaporize into steam, starting the actual boiling process. And other small portions of water would be heated sufficiently to turn to little steam bubbles that would also rise to the top and each little bubble would 'violently' turn to steam as it was released.

    Yes, just like this oven, the microwave oven also heats evenly, and the lack of swirling water of a pan on the stove top is definitely a factor in superheating. But the real magic is the bubble formation AND the release of a bubble or two to the disturb the surface, then you have boiling! You might wonder about those casseroles you pull bubbling from the oven, why don't they get superheated? It's because they contain chunks of 'foreign' objects, like meat and vegetables that cause uneven heating. If you have water or 'coffee-water' or even very well-stirred cocoa, you can get superheating under the right conditions. Be careful!

    And don't start with me about that crap saying injuries from superheating are just urban myths! Using only the intelligence found in ant feces, you will realize why more superheat incidents aren't well reported. A person reaches in the microwave for their coffee, it explodes burning their arm, and if it's severe enough, they go to the hospital emergency room for treatment. Or perhaps someone is boiling water on the stove making spaghetti, and as they pour it into the colandar in the sink, they slip, pouring the water much too fast, it splashes back on them, severely burning their arm. The doctor reports both incidents the same - 'cooking accident - burned with boiling water'. Even if the patient starts with the microwave story or adds a routine about aliens bursting in while they were cooking and burned them by peeing on them with boiling water, the doctor's report will still be the same, 'cooking accident - burned with boiling water'. Got that?

    You, a Terrible Chill, and Your Electric Blanket

    When some people are sick, they get fever and chills and sometimes the chills are so severe they produce almost uncontrollable shaking. Some people resort to an electric blanket, with its control set very high, to wait for the fever to 'break'.

    How dangerous or ill-advised this action is, can only be answered by a medical professional. The following is just another example of some data that might be interesting.

    The formula to determine the Power of Your Microwave Oven is used for this problem as well. The question: if a person lays under an electric blanket with its thermostat set very high, what temperature rise will the person's body experience?

    The formula, restated from above:
    Power = (gal. X temp rise X 8797) / seconds
    Except we want to solve for temp rise, so the formula becomes:
    temp rise = Power X seconds / (8797 X gal.)

    In this case, the body weight is 150 lb. Because each gallon of water weighs 8.34 lbs. (and the body is primarily water), 150 lb = approximately 18 gallons. The time is one hour or 3600 seconds. The power of the electric blanket is usually found on the bottom of the control, in this case, 180 watts.
    Now plugging the numbers in, temp rise = 180 X 3600 / (8797 X 18) = about 4° F!

    In other words, a 150 lb. person laying 1 hour under a 180 watt electric blanket turned full up, might experience a 4° F. temperature rise.

    Warnings and Caveats
    Naturally this assumes the blanket control is constantly on - it may cycle on and off, even though it's turned to the maximum position. This would produce a correspondingly lower result. Additionally the 'weight value' doesn't allow for the weight of the head that is usually not heated, causing at least a 5% error in the other direction. And only a portion of the blanket makes contact with the body, so not all the blanket's power is transferred to the body. Because the body is not pure water, it will not react as though it was, further degrading the formula. As you can see, this may only be a vague approximation.

    Check with your doctor before trying this.

    Two 110v Incandescent bulbs, a 25 Watt & 100 Watt, in series.
    Which is brighter?

    This is almost a trivia puzzle, but illustrates some very real and useful properties of incandescent bulbs.

    The puzzle is quite simple - if you connect TWO 110 volt, different wattage, incandescent bulbs IN SERIES, which will burn brighter?

    For the non-technical readers, the term "in series" means the bulbs are connected 'end to end' and then to the 110 power - rather than "in parallel", where each bulb is connected directly across the power. (Two flashlight batteries are connected "in series", two different tablelamps are plugged "in parallel".)

    If you have any knowledge of resistance and Ohm's Law, you might as well make your poor guess now. For unless you KNOW the answer from previous experience, you're likely to be in the 90 percentile of technical people who guess incorrectly.

    The puzzle is made more difficult by someone using more thought and pulling more trivial data into the mix. Each piece of data seems to lead the answer in opposite directions.

    The setup and rules are simple, two different wattage bulbs are connected in series; which one is brighter? The term 'brighter' can be defined in most any manner you choose. It can be 'visually brighter', it can be the amount of infrared energy radiated, or the most voltage across each bulb may be the criteria. It's NOT a trick question involving semantics. The bulbs are standard incandescent bulbs, not 'funny' bulbs, although you could use various types of similar bulbs, like flood lamps and chandelier bulbs. And the exact wattages aren't important, except that 100 Watt and 25 Watt are needed for the math explanation that follows.

    (If the bulbs are the same wattage, they will be equally bright - UNLESS one is shorted, then one will be dark and the other fully bright. If one bulb is open, then still, as before, they BOTH will be 'equally illuminated'... )

    So, WHICH is brighter? 100 Watt or 25 Watt Bulb?

    (Scroll down for the answer)
    It's the 25 Watt bulb.

    BUT the reasons why are far more interesting than the answer!

    First, measure the resistance of the bulbs. The 100 Watt bulb measures about 10 ohms and the 25 Watt bulb is about 40 ohms. Now add those two together and we get 50 ohms, meaning about 4/5ths (40 / 50ths) of the total resistance is in the 25 Watt bulb, and it drops 4/5ths the total voltage, leaving only 1/5th for the 100 Watt bulb. If the 25 Watt bulb operates on 4/5ths the normal voltage, it will dissipate about 20 Watts. While the 100 Watt bulb is operating with only 1/5th the normal voltage and dissipates about 1/5th 100 Watts or 20 Watts.

    We just proved the 100 Watt bulb consumes 20 Watts while the 25 Watt bulb consumes 20 Watts!!

    Correction: The new answer is, BOTH bulbs are equal. (??)

    Isn't it?

    If NOT, it must be a silly mistake in the math! Where'd we go wrong?

    OK, from the top. The 100 Watt bulb measured 10 ohms, let's see if that measurement was reasonable. The formula we want is P = E^2 / R. That is take the voltage (E), square it, divide by the resistance (R) and you get power (P). So 110^2 = 12100 / 10 = 1210 Watts.

    Hmmmm. 1210 Watts is NOT what a 100 Watt bulb dissipates!!! THAT surely qualifies as a math error! How could our resistance measurement have been off by 10 TIMES, or about 1000% wrong?

    That problem is addressed by a characteristic of incandescent bulbs - their resistance VARIES! It has a positive temperature coefficient. As the temperature goes up, its resistance also goes up. So as the bulb's filament rises from room temperature to about 4600°, its resistance increases about 10-fold, very conveniently, for our calculations.

    The '10 ohm' 100 Watt incandescent bulb is really a 100 ohm 100 Watt bulb while it's operating. Likewise the '40 ohm' cold 25 Watt bulb is really a 400 ohm bulb while it's on. (Notice the ratio of resistance remains precisely the SAME as before, 4/5ths and 1/5th.)

    NOW let's see if the calculations behave better with this new data. We have 100 ohms plus 400 ohms, totaling 500 ohms. Using the E^2 / R = P formula, we have 110^2 = 12100 / 500 = 24 Watts total used by the two bulbs in series. Still with 4/5ths the resistance in the 25 Watt bulb (400 ohms of the 500 total), we determine about 4/5ths 24 Watts or roughly 20 Watts goes to that bulb. And that leaves ONLY 4 Watts (24 Watts total - 20 Watts = 4 Watts) left for the 100 Watt bulb.

    Viola! Now the numbers work! The 25 Watt bulb gets 20 Watts, nearly a full load, while the 100 Watt bulb only gets 4 Watts.

    If both bulbs increase their resistance 10 times when operating, what was wrong with our calculations using the COLD values?

    The mistake was assuming the TOTAL Wattage drawn by the combination of bulbs. The first calculations ignored the total effective Wattage really drawn by the combination of both bulbs. And we just used the resistance RATIO of the bulbs, applying that ratio to the rated Wattage of the bulbs. By ignoring the TRUE total Wattage of ONLY 24 Watts, we used the FULL rated Wattage of the 100 Watt bulb to arrive at 1/5 its FULL Wattage, 20 Watts, rather than 1/5 of the actual 24 Watts drawn, or the correct answer of about 4 Watts.

    So, we're done?

    Nope, the errors are still substantial. Remember how the resistance of incandescent filaments change with temperature? (It's the basis for this whole mess.) If the 25 Watt bulb is operating at 4/5ths the normal power, its resistance should be near normal. But what about the 100 Watt bulb operating at 1/5th normal? It's barely on, so its resistance, rather than around the normal 'hot' 100 ohms, is MUCH closer to the 'cold' 10 ohms.

    This little detail changes our calculations even further, where the 25 Watt bulb is almost operating at full power, its resistance is very close to 400 ohms (hot) yet the (cold) 100 Watt bulb is still only about 10 ohms. And for a final answer, we have only about 2% of the total power going to the 100 Watt bulb and 98% to the 25 Watt bulb.

    Clearly the 25 Watt bulb is very bright and the 100 Watt bulb isn't producing ANY light!

    The 25 Watt bulb WINS!

    Warnings and Caveats

    If you build an example of this to verify these results, exercise due caution!! Remember, any contact with 110 volts can be deadly!
    Do NOT try this if you are unfamiliar with electrical safety measures.

    Final thoughts: the combination of a 100 Watt bulb and a 60 Watt bulb acts a bit different, as the bulbs are more closely matched and when first connected, the 100 Watt bulb pulses bright for an instant. But as the bulbs stabilize, again the 100 Watt bulb loses, although this time it's allowed to slightly glow...

    If the 100 Watt bulb is about 100 ohms in normal operation, producing 100 Watts of light (actually 10 watts of VISIBLE light), how do you think it acts during the first few instants after the power is applied and it's still about 10 ohms? Yep, it's effectively a 1000 Watt bulb - and THAT'S why bulbs tend to burn out when you first turn them on...

    Home      Information      Photos      Download Files      Links      Site Index     

    Copyright © 2003- Tom Scogin    Norcross, GA.    All rights reserved.