So I just boiled my kettle (500ml, fresh from tap). Mains here = 242
volts, Kettle takes 12.3 A (pretty much constant) Power taken = 0.055kWh
(so pretty close to my calculation!) I allowed the kettle to trip using
the steam generated by the then boiling water (so addressing both the
loss and latent heat aspects of the question.)
On 29/08/2022 12:28, Peter Merchant via dorset wrote:
I had considered that, but it's so long ago in my ancient past that I
studied thermodynamics, I didn't look into calculating it. My concern
about it is the cost of raising the temperature to 100 deg, then the
latent heat of vaporisation to actually boil it.
Peter M. [ Applies to any form of boiling water of course. ]
n 29/08/2022 11:31, Ian Morris via dorset wrote:
500ml of water = 500g water. let's assume you want to go from 20-100
C => increase of 80C
Specific heat capacity of water = 4.184 J/g-K
=> 167,360 J = 0.0465 kWh Electricity @27p/kWh = 1.2552p
So rather similar (OK, I don't know efficiency of electric kettle
... but heating elements are 100% efficient... sure there will be
some losses to both the kettle and the wider environment, but i doubt
somehow that they are huge) Of course this is about cost for the
consumer ... I'm totally ignoring the ineffeciency of turning gas
into electricity and transmitting it around the country in the first
place.
On 29/08/2022 09:33, Peter Merchant wrote:
I was curious, so I took the water from the kettle and put it in a
pot on the stove. It was very close to 500ml.
Watching the gas meter while it boiled it used 0.014 m(cubed) of gas,
which converted to 0.15788Kwh and at my current rate of 7.123p/Kwh
cost me 1.1.25p to boil up.
Unfortunately I can't see my electricity meter easily to see what it
costs me to boil that amount in the kettle or by Microwave. Does
anybody have that facility? And also there is always other
electricity being used.
Cheers,
Peter
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