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25-01-2015, 16:27 | #1 |
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Heavy science Question?
My wife and I had an interesting conversation this morning, to which neither of us know the answer. I thought a quick look on Google would answer it, but I can't find it on there either...
So here is the question... Assuming there are no outside losses, the room is thermostatically set at 20c. I have 2 identical jugs of water each with exactly 1 litre of water in, but one is at 30c, and one is at 10c. if both jugs are left in the room, which will get to 20c first, the hot or cold one? Also, along the same lines, if I had 1/2litre at 90c, and poured 1/2litre of 10c water into it, what would be the end temp of the 1litre. To me, these should be a fairly easy way to work this out, but I can't find it anywhere. |
25-01-2015, 17:14 | #2 |
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I know a guy who knows a guy. Let me get back to you
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25-01-2015, 17:52 | #3 |
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http://en.m.wikipedia.org/wiki/Mpemba_effect
Check this out see if it helps, and this one maybe...... http://sciencefair.math.iit.edu/projects/cooling/ |
25-01-2015, 18:51 | #4 |
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As the colorific content of the 30C jug is higher I would assumethe 10C jug would get there first
The higher temp one would have more to lose than the lower one had to gain |
25-01-2015, 19:10 | #5 |
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Ok answer from someone I know with a good scientific background.
"Pretty straight forward for both really. For the first one. The 30' jug will cool down more rapidly than the 10' jug because it has a greater temperature difference from the room, so the temperatures will converge, BUT. it will never quite catch up with the 10' jug, so the 10' jug will reach room temp first, but only just. The mixing of the two is simply an average. (90 + 10) / 2 = 50'C You can work it out by more complicated means and calculate the amount of energy contained in the two jugs and then work out what temperature that would make 1l of water, but you still get the same answer." Clever stuff I guess |
25-01-2015, 19:15 | #6 |
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So I woz right
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25-01-2015, 20:25 | #7 | ||
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Quote:
Quote:
The second one makes sense, when I used to fill the Fish tank, it would take about 4l of boiling water to bring the 16l of 10' tap water up to 30', which I guess is about right. I am surprised how hard this is to find the answer too. |
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25-01-2015, 21:49 | #8 |
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just seen this, and my answer would have been, you always lose heat quicker than gain it so in theory the hotter one would get to 20 first but as they near to the target temperature the hotter one will slow down so I would say they will take almost the same time, Rick
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25-01-2015, 22:12 | #9 | |
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Finally found a way to word the question, where Google has come up with something productive... First I found this site... it reckons heat is symmetrical, so they would be identical...
http://askascientist.co.uk/physics/d...an-warming-up/ I carried on looking, and basically, as long as the elements are all identical, eg, both cooling or heating the same amount in the same ambient temp, then they will be identical. The main thing is, heat is given off, or taken on in a way where the bigger the gap, the higher the initial heat transfer, which then reduces as they get closer to ambient temperature... this leads to this interesting anomaly... Quote:
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25-01-2015, 23:11 | #10 |
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I picked him up on the temperature difference
He responded with this? If any Facebook cloners currently reading this thread would kindly refrain from cloning my personal details and or friends list |
25-01-2015, 23:19 | #11 |
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at the end of the day they will stabilise at the same time, Rick
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25-01-2015, 23:24 | #12 |
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25-01-2015, 23:28 | #13 |
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what
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25-01-2015, 23:30 | #14 |
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