And on your way to the gas meter you can kill the electricity at the consumer unit. No need to reach across the conflagration whatsoever.
Electric socket too close to the 'hot zone'?
- Thread starter Jamestiger150
- Start date
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That would certainly be sensible to avoid things becoming even worse, but once the appliance was a 'blazing inferno' it could well continue to be even in the absence of a gas supply.
Once a fire is established, operating any electrical switch will achieve nothing, other than preventing subsequent electrical problems due to fire damage of cable downstream of the switch (if not already affected by the fire).
I think this discussion may be getting a little silly. Although I wouldn't design it as such, and despite AO's rules (which are their prerogative), I would be perfectly happy to live with the situation depicted in the OP (with no expectation of ever operating the socket's switch). Better, I might move the socket down to below counter level, such that I would have no switch to operate, or retain a switch in that position, feeding an unswitched socket below counter level.
Kind Regards, John
+1
Sorry John, but physics is not an opinion.
Conductivity between the water and the pan dictates that as the water is at 100⁰C, the pan will be at least at that same temperature.
Water at 100⁰C is in direct contact with the pan an all sides, hence, any part of the pan will be at least 100⁰C.
However, as metal is a very good conductive material, as it is heated at the bottom by a flame (~2000⁰C) the sides of the pan will also get hot, not as much as the bottom because of dissipation, but surely more than 100⁰C.
In fact, if you left the pan full of water on the gas, the water will evaporate, full proof that it's been heated beyond the 100⁰C.
I hope it's clear.
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Indeed, at least at this level (get onto particle physics / astrophysics / quantum mechanics etc. and "opinions will vary"
). Traditional physics was not a matter of opinion when I was at uni, and still isn't [mind you (our understanding of) even 'traditional physics' is not set in stone, as the likes of Einstein have demonstrated - and, of course, Physics once 'said' that everything orbited around planet Earth! ]! Anyway ....I would say not "at least" for all parts of the pan - that will only be true for the parts of the pan through which heat is travelling into the pan - and, in fact, since metal is a very good conductor, even for the base, the outside temp will not be much above the ~100°C temp of the inside surface of the base . For other parts of the pan (i.e. the 'sides'), where heat is travelling out of the pan into the surrounding air (IF you accept that happens - see below), the temp gradient is in the opposite direction, so the outside of the pan will be (again, probably only fractionally, due to the good conductance of metal) below (not 'at least") the ~100°C temp of the inside surface of the sides.
Ah - are you saying that you don't believe that heat from within the pan will 'lost' from the sides of the pan into the surrounding air?
Whatever, quite part from theory, we have the empirical evidence from my experiment this afternoon to consider. I was able to touch, for a few seconds, the outside of a kettle in which water was being boiled - and I think I would now 'have the scars' had that been "...surely more than 100⁰C" ?
Once all the water (which was the only thing 'capping' inside-pan temps at ~100°C) has gone, the goalposts will obviously have totally moved, and all of the pan will try to rise to a temp approaching that of the source (flame or whatever), failing to completely achieve that goal only because of heat loss into the surrounding air etc..
Kind Regards, John
It doesn't work like that.Indeed, at least at this level (get onto particle physics / astrophysics / quantum mechanics etc. and "opinions will vary"
I would say not "at least" for all parts of the pan - that will only be true for the parts of the pan through which heat is travelling into the pan - and, in fact, since metal is a very good conductor, even for the base, the outside temp will not be much above the ~100°C temp of the inside surface of the base . For other parts of the pan (i.e. the 'sides'), where heat is travelling out of the pan into the surrounding air (IF you accept that happens - see below), the temp gradient is in the opposite direction, so the outside of the pan will be (again, probably only fractionally, due to the good conductance of metal) below (not 'at least") the ~100°C temp of the inside surface of the sides.
Ah - are you saying that you don't believe that heat from within the pan will 'lost' from the sides of the pan into the surrounding air?
Whatever, quite part from theory, we have the empirical evidence from my experiment this afternoon to consider. I was able to touch, for a few seconds, the outside of a kettle in which water was being boiled - and I think I would now 'have the scars' had that been "...surely more than 100⁰C" ?
Once all the water (which was the only thing 'capping' inside-pan temps at ~100°C) has gone, the goalposts will obviously have totally moved, and all of the pan will try to rise to a temp approaching that of the source (flame or whatever), failing to completely achieve that goal only because of heat loss into the surrounding air etc..
Kind Regards, John
The side of the pan in contact with water will be at 100⁰C +.
Heat dissipation through contact with air won't be enough to drop its temperature to remarkable levels because heat (energy) is applied to the system by the flame and transferred from the bottom of the pan to the water and side of the pan as well as directly from the bottom to the sides. (Second principle of thermodynamics)
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