GR9677 #13



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benjamin_DW 20161027 14:18:09  What\'s not clear to me is why the substance **must** cool at the same rate that we heated it. Just because we heated it at a rate of 100W does not mean it will necessarily cool at a rate of 100W. In fact, by Newton\'s law of cooling, the temperature is given by: \r\n\r\n\r\n\r\nWhere is the background temperature, and is a constant dependent on the geometry of the material.[and even this of course, is an approximation]
rweads 20170430 01:28:20 
since the liquid does not boil, and the heat element has been in the water for a long time, the system must be in thermal equilibrium where the rate of heat added is equal to the rate that heat is dissipated. It is then very reasonable that for small deviations from that equilibrium temperature the rate of heat loss will be the same.

  cczako 20131017 17:07:38  The easiest way to do this is to use units. 4200 J/ kg C times 1 C and 1 kg (1L=1kg) is 4200 J. Watts is Joules per second so divide 4200 J by 100 J/s and you get 42 s.   asafparis 20090402 13:10:33  of water is not
shak 20100814 09:51:27 
1 L of water is exactly one cubic meter!
m= v*r

BerkeleyEric 20100921 19:13:11 
1 liter is equal to 0.001 cubic meters, which does correspond to 1 kg for water

Rtrt 20151013 08:49:28 
yes . 1L is 1 kg but it is not 1m^3. density of water is 1000 kg/m^3. 1 L is 0.001 m^3. So mass is 1 kg.

  naama99 20061121 08:18:36  Since the formula for specific heat c has delta T in it, why do you take 1 degree as delta T instead of 99?
welshmj 20070731 14:37:33 
because the problem says that it cools by 1 degree not that it cools TO 1 degree which means that the change in temperature (delta T) is 1 degree not 99 degrees.

  vortex 20051204 16:09:10  not ,and an arrow would be better before 1 kg.   yosun 20051110 23:32:23  angiep: the specific heat of water is 4200 J/kg and not 2200. thanks for the typoalert.   angiep 20051110 23:14:50  I dont feel like this problem makes any sense at all. It is worded completely wrong.
einstein 20060331 02:32:57 
In fact it is worded perfectly correctly. The statement that the 100Watt heating element remains for a long time but the water does not boil informs us that the water must be disipating the energy at precisely this rate. Hence P_{out} =100W

Imperate 20080904 09:02:50 
I can see it makes sense now thanks to einstein's comment. However I have to agree that this question is worded very badly. The sentance "the heating element is on for a long time and the water although close to boiling does not boil" makes it sound like the water heats up to being close to boiling, not that the water started close to being at boiling point and despite the heaters presence,did not boil.rnWould have been much better to say the temperature of the water remains static, despite heater of 100W being placed in it.

tachyon788 20091006 09:58:33 
I agree with Imperate. I thought the question meant that the heater heated the water (and increased the temperature) until it was removed right before the water started to boil. In that case, the whole heater thing seemed like a red herring. With only 1.7 minutes per question, we don't really have time to decipher poorly worded questions...

hdcase 20091101 22:06:40 
I disagree with Imperiate. The question isn't saying that the water started at close to boiling  rather, it is saying that the steady state of the system is such that at the steady state temperature (which is close to 100 C) the system is losing 100 W. I think it is worded in such a way that one can glean that information without it having to be stated directly.

 

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