Seems to me that the more power you apply in an electric kettle, the
Your time obviously isn't worth much to you. Boiling a kettle uses only a couple of cents' worth of electricity. A teapot with a cozy on it, or a thermos jug, keeps it hot for as long as you're likely to want it, no?
Cheers
Phil Hobbs
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Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
A tea ball is your friend. Just fish it out after 5 minutes or so.
Personally I just strain out the tea leaves with my teeth. ;)
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
Not only that but in the US the power is limited to ~1700W that together with modern toasters catering for "thick" bread, or 4 slices, or being a combined toaster oven means that the heat flux is too low to toast effectively.
The bread is dried out before it toasts.
That's not so much of a problem when you can get 3kW out of a wall plug.
I had an old Proctor Silex from 30 years ago that worked well, but my wife threw it out in favour of a more modern Cuisinart that was terrible. I still haven't found a modern one that I like.
Not hard to estimate. A spherical tea-kettle of radius R has a volume of 4/3 pi R**3 and a surface area of 4 pi R**2. We assume the contents are well stirred, so that the water is isothermal.
With a wall of thickness d (assumed small compared with R) made from a material of thermal conductivity alpha, a temperature drop of T across the wall will pump
Q dot = 4 pi R**2 T/(d alpha)
watts into the environment. The cooling time constant is then
tau = Q dot / (V rho c_P T)
where rho is the mass density and c_P is the specific heat. For water, rho is 1000 kg/m**3 and c_P is 4.2 kJ/kg/K.
Thus the cooling time for the thermally grounded kettle would be
(d alpha) (4/3 pi R**3 * 1000 * 4200 * T) tau - ------------------------------------- (4 pi R**2) T
tau = 4200000 d alpha R -------------- 3
For an 8-cm radius kettle (about 2 litres), a 4-mm wall thickness (probably about right for glass), a thermally grounded glass kettle with alpha = 1 /W/m/K would have a time constant of
tau = 4200 * 0.004 * 1000 * 0.08 / 3 = 480 seconds, or 8 minutes.
With aluminum, whose alpha is about 200 W/m/K, it would be 2.4 seconds.
The reality is probably about 15 minutes as a SWAG. So the glass kettle walls actually help significantly.
Of course if you're heating it on a stove, it works against you there.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
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It's the chemicals in NY water. Did you see the recent news item about NY water having unsafe chemicals ?>:-} ...Jim Thompson
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You must have more sensitive taste buds than I have. What tea brands do you use. My wife buys PC Tips from some internet store. Not the best tea buy better than a lot of teas.
I believe when you boil it the first time you drive out all the dissolved oxygen. And when you boil it again, it is the same as when only boiled once. We use an electric kettle, so it does not get boiled for a long time.
That's the best part. Then you add the East River sludge >:-} ...Jim Thompson
-- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | San Tan Valley, AZ 85142 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at
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