This is the trick question I was asked many many years ago when I was but an impressionable young man hard at my studies.
To all intents and purposes the kettle will indeed return to room temperature - but here lies the trick - if you measure the temperature of each and every molecule of water in the kettle then the chances of them all being the same as room temperature is impossible, some will have a higher temperature, and some will be lower.
So the answer is no!
Note that I have used 'temperature' here - this should really read the 'enthalpy' of the molecules.
You can work out the probability if you wish - 1 mol of water (18g or 18ml) contains 6.022 x 1023 molecules.
2006-12-31 02:59:14
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answer #1
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answered by Anonymous
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I think your tutor was talking about the fact that the rate of transfer of heat is related to the temperature difference, so when the temp diff becomes less and less the rate of heat loss also falls so it never quite gets to room temperature (in a perfect situation where the room temperature is steady). Actually "room temperature" is always going up or down a bit and varies slightly from place to place, so before very long the temperature of the kettle will get down to the range of temperatures the room is covering. Then you can say it is in equilibrium and as much at room temperature as anything else in the room.
Practical answer not a mathematical one.
The stuff about planck temperatures and numbers is not really relevant to large scale things like rooms and kettles. It just means the behaviour is not absolutely smooth like a mathematical function when you get very very very tiny.
2006-12-30 17:31:00
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answer #2
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answered by philjtoh 2
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Do this thought experiment:
Boil the kettle, and then place it in a deep-freezer.
On its way from boiling to sub-zero, the temperature function is continuous. Therefore, the temperature *must* reach room temperature at some point on the way down after boiling.
By basic thermodynamics, if the boiled kettle is placed in a room temperature environment that is sufficiently large and open, it will reach equilibrium at room temperature.
Where this gets tricky is when the room is small and closed, and the heat from the kettle re-sets the room temperature to an equilibrium temperature that is higher than the original temperature. Nevertheless, the ultimate equilibrium temperature will be exactly equal to the "new" room temperature.
2006-12-30 14:47:01
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answer #3
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answered by Jerry P 6
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There is a Planck's constant in quantum mechanics. Its value is 6.63 Ã 10â34 J-sec. If you multiply the frequency of a photon by Planck's constant, the result is the energy contained in the photon. If your kettle is left in the room long enough it will reach equilibrium with the air in the room and its temperature will not be measureably different from the room's air temp.
2006-12-30 15:03:26
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answer #4
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answered by curious george 5
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yes it will.your tutor is completely wrong.... temp of boiling water is 100degrees C..heat is transferred from a hot body to a cold body...so the heat is lost to the room as the room temperature is normally less than the boiling point of water.this process will continue until the temperature of water and the room temperature become equal.therefore water will cool down to the room temperature...the room temperature might rise a bit but it will equalize with the temperature of water
2006-12-31 13:30:35
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answer #5
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answered by nipun 2
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Exactly what is room temperature? What ever the temperature in a room is,that is room temperature.And, yes it will eventually return to the temperature that is in that room.
2006-12-30 14:51:10
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answer #6
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answered by Anonymous
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I dont really get the question but yes it will go back to room temperature again it just may take time to cool down again.
2006-12-30 14:45:30
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answer #7
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answered by susank2000 1
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He's right. Space, time and temperature all have minimum units. Nothing is infinitely divisible.
2006-12-30 17:58:45
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answer #8
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answered by Nomadd 7
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of course
2006-12-30 15:07:37
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answer #9
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answered by I am not afraid!! 1
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