Water heating experiment

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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
130921-1146 EDT

A couple years ago I compared heating water in a Hot Pot vs a Microwave. Today I tried this experiment with a Silex type container on a single burner electric element.

I have calculated that theoretically about 1.095 wH are required to heat 1 quart of tap water 1 degree C in the range above freezing and below boiling. Someone check my figure. My reference was the 40th edition of the Handbook of Chemistry and Physics, p 2107.

Using a Rival 4071WN hot pot and 1 quart (32 ounces) of tap water I measured an efficiency of 1.095/1.153 = 95%.

Using a GE microwave with the water in a 1 quart Pyrex measuring bowl the efficiency was 1.095/2.6 = 42%.

Today's test, the equivalent of a glass pot on an ordinary electric range top, showed an efficiency of 1.095/2.12 = 52%.

What happens when the test is performed with a cast iron pot on an induction heating range? One problem here is the large metal mass in relationship to the water mass.

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mgookin

Senior Member
Location
Fort Myers, FL
130921-1146 EDT

A couple years ago I compared heating water in a Hot Pot vs a Microwave. Today I tried this experiment with a Silex type container on a single burner electric element.

I have calculated that theoretically about 1.095 wH are required to heat 1 quart of tap water 1 degree C in the range above freezing and below boiling. Someone check my figure. My reference was the 40th edition of the Handbook of Chemistry and Physics, p 2107.

Using a Rival 4071WN hot pot and 1 quart (32 ounces) of tap water I measured an efficiency of 1.095/1.153 = 95%.

Using a GE microwave with the water in a 1 quart Pyrex measuring bowl the efficiency was 1.095/2.6 = 42%.

Today's test, the equivalent of a glass pot on an ordinary electric range top, showed an efficiency of 1.095/2.12 = 52%.

What happens when the test is performed with a cast iron pot on an induction heating range? One problem here is the large metal mass in relationship to the water mass.

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It takes 4.187 Joules to raise 1 gram of water 1 degree Celcius.
1 Joules = 1 watt for 1 second.
There are 3600 seconds in an hour.
1.095 wH = 3942 Joules
Density of water = 1 gram/ ml
32 oz = 946 ml.

To raise 946 ml of water one degree celcius should take 946 * 4.187 = 3960 Joules.
To convert that to wH it's 3960/3600 = 1.100 wH

Looks like we're pretty close to each other.
 

mgookin

Senior Member
Location
Fort Myers, FL
I'm a little suspicious of that hot pot test though. Could you describe the test method please?
Is it preheated?
Is there anything stirring the water to provide for more uniform distribution of the heat?
How are you recording temperature of the water? A submerged probe?
I'm assuming volume and pressure are not constrained. I'm not familiar with a hot pot; is that more like a crock pot or a pressure cooker?

What happens if you scale it up and elevate the temperature 20 degrees?
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
130921-1439 EDT

mgookin:

Thanks. I wanted verification I did not make a mistake.

The values I used were:
Density of water at 75 F (28 C) is about 942.8 grams/quart from 3771 grams/gallon.
Specific heat of water 4.1804 joule/gram/deg C.
1 joule = 1 watt-second.
3941 watt-seconds/quart/deg C. 1.09472 watt-hour/quart/deg C from my values.

But test measurements on heating water might have a 10% or so error.

Your density of water was a 0 deg C and mine at 28 deg C.


To your next post:

The hot pot is a one quart size aluminum container with the 1 kW heating element protruding into the water. For these tests the thermostat was bypassed by being set to maximum. Water was raised from 75 F to 200 F. Time about 5 minutes. A mercury thermometer was used for temperature measurement. Self stirring, but thermometer was used for a little.

The biggest error is about 10 % resulting from kWh resolution of 0.010 . But some energy goes into the container, evaporation, and conduction and radiation. This was not an accurate test, but one to show the big difference between different means of heating water.


No pressure constraint, just open to the air, no cover.

The microwave test has errors of a thick Pyrex measuring container, and the 10 Wh resolution.

I am not sure of your question:
What happens if you scale it up and elevate the temperature 20 degrees?
Anywhere within 10 to 90 C is probably OK. But need finer resolution on energy measurement.

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GoldDigger

Moderator
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Location
Placerville, CA, USA
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Retired PV System Designer
That clarifies it. From your OP I thought you were only raising the temp 1 degree.

So this warmer has a submerged resistive heating element and you're measuring 95% efficiency?

That sounds about right as long as the hot pot itself has low thermal mass and is reasonably well insulated. A tall narrow pot will also lose less heat from the water surface to the air while heating than a wide shallow pan or pot if both are uncovered.
By comparison, the microwave has losses in converting electricity to microwave energy and a conventional stove top heats up a lot more then the pot of water.

BTW, once you get to the point where the water is boiling, covering the container becomes a lot more important, since boiling carries off heat really fast.
 
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mgookin

Senior Member
Location
Fort Myers, FL
Very good test indeed. Too bad the grid is only 50% efficient by the time the power gets to our homes & businesses. That grid won't exist one day. I envision little power plants about the size of our a/c systems at each site one day. If not that first, we should at least have localized power generation in each community. And why don't they have a Faraday cage around transmission conductors? Wouldn't that reduce waste?
 

mivey

Senior Member
Using a Rival 4071WN hot pot and 1 quart (32 ounces) of tap water I measured an efficiency of 1.095/1.153 = 95%.

Using a GE microwave with the water in a 1 quart Pyrex measuring bowl the efficiency was 1.095/2.6 = 42%.

Today's test, the equivalent of a glass pot on an ordinary electric range top, showed an efficiency of 1.095/2.12 = 52%.
I suspect there are a lot of variables. I can heat up a bowl of soup faster on the stove than I can in the microwave, but I can warm a leftover steak in the microwave faster than I can in a frying pan.
 

mivey

Senior Member
Too bad the grid is only 50% efficient by the time the power gets to our homes & businesses.
The grid is closer to 85-90% efficient. About 4-6% loss from generator to substation. About 5-10% from substation to homes and businesses. As for local loss comparisons: local small generation (at least that similar to existing technology) won't be as efficient as the larger units and neither will fuel delivery.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
130922-1244 EDT

The purpose of the experiments was to provide examples of measurements that the ordinary electrical customer could perform with a Kill-A-Watt and show substantial energy differences for the same task.

This does not mean that hot pot should be used to heat water instead of the microwave. In my home we use the gas stove to heat water, faster and cheaper. In our area the cost for thermal energy from gas is between 1/3 and 1/7 of that for electricity, excluding heat pumps. The 1/7 can not be achieved because of the inefficiency of gas compared to 100% for electric.

If I needed a cup of hot water I would use the microwave.

An additional experiment may come in the future by way of K8MHZ.

With the advent of Smart Meters it is possible for the average consumer to do some energy and power measurements that may help them reduce their energy use without purchasing additional instrumentation. However, at about $30 for the Kill-A-Watt EZ meter the average consumer could make experiments on some products easier than trying to use the Smart Meter. Also many libraries are loaning Kill-A-Watt meters.

The 10 Wh resolution of the Kill-A-Watt produces a substantial possible error in the hot pot measurement, but that does not deny the result that the microwave is a lot less efficient than the hot pot.

To maintain 1 quart of water at 180 deg F in the hot pot requires about 100 W. Or 0.1 kWh per hour of operation, $0.016 per hour for me. Thus, in 5 minutes 0.0083 kWh of energy is output thru the container.

During the heating experiment it is reasonable to estimate 1/2 of this loss was part of the energy to raise the water to the 200 deg F level. In the hot pot test 0.08 kWh was used and subtracting 0.0042 the result is 0.076 kWh or the theoretical requirement. But there is still the possible 0.01 kWh error. And we know that energy went into the mass of the container. Much better equipment is required to do a better test. But that was not the object of the experiments.

I prefer a gas range top for kitchen purposes, including melting glass. I use the kitchen for various experiments. In times long past my dad used the gas stove to melt lead or babbit in a small cast iron ladle.

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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
130922-1532 EDT

My dad also heated sulfur on the kitchen stove. That was smelly. Whether my mother objected to lead I don't know, but possibly the sulfur.

What really caused an objection was when my grandfather brought home a skunk and wanted to cook the oil out in my grandmother's oven.

When Henry Ford made his first gasoline engine, 1893, it was first run on the kitchen sink. His wife dripped gasoline into it, and ignition was from a wire to the light socket. http://blog.thehenryford.org/2013/09/everything-but-the-kitchen-sink-engine/ Henry started at Edison in 1891 and became chief engineer in 1893. Henry had a telephone at his Bagley home, and was available any time day or night to solve problems at the power plant. http://corporate.ford.com/our-company/heritage/heritage-news-detail/650-henry-ford


More on kitchen experiments:
http://www.goodyear.com/corporate/history/history_story.html

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