B
bz
Guest
Jasen Betts <jasen@free.net.nz> wrote in
news:f6st4t$832$1@jasen.is-a-geek.org:
1 cup is about 236.6 ml.
It take 1 cal to raise 1 ml of water by 1 degree.
A watt is about 0.239 cal/second.
Let us assume the cup of water starts at 25 C and you want to raise it to
100 C, that is 75 degrees that we want to heat the water.
So, it will require (236.6 ml x 75 deg x 1 cal/(deg x ml)) = 17744.25 cal
to raise a cup of water 75 deg C.
You can deliver those cals at any rate you like[see notes below]. A 1 kw
heater will deliver 239 cal per second. A 2 kw heater will deliver at
twice that rate.
74.3 seconds to heat a cup with a 1 kw heater.
37.1 seconds to heat a cup with a 2 kw heater.
[notes: ignoring losses in wiring, assuming efficient heat transfer from
heater to water, assuming no loss of heat from cup+heater combination]
If you want to factor in any of those, please state your assumptions [such
as thermal transfer resistance between heater and water, between container
and air. Those act much like series resistors as far as the heat is
concerned. You can model temperature difference as voltage drop and heat
transfer as current if you like. Any good engineering heat transfer text
will show you exactly how to do this{but in this case, it won't really
matter all that much}].
--
bz 73 de N5BZ k
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.
bz+ser@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
news:f6st4t$832$1@jasen.is-a-geek.org:
Some useful information:
1 cup is about 236.6 ml.
It take 1 cal to raise 1 ml of water by 1 degree.
A watt is about 0.239 cal/second.
Let us assume the cup of water starts at 25 C and you want to raise it to
100 C, that is 75 degrees that we want to heat the water.
So, it will require (236.6 ml x 75 deg x 1 cal/(deg x ml)) = 17744.25 cal
to raise a cup of water 75 deg C.
You can deliver those cals at any rate you like[see notes below]. A 1 kw
heater will deliver 239 cal per second. A 2 kw heater will deliver at
twice that rate.
74.3 seconds to heat a cup with a 1 kw heater.
37.1 seconds to heat a cup with a 2 kw heater.
[notes: ignoring losses in wiring, assuming efficient heat transfer from
heater to water, assuming no loss of heat from cup+heater combination]
If you want to factor in any of those, please state your assumptions [such
as thermal transfer resistance between heater and water, between container
and air. Those act much like series resistors as far as the heat is
concerned. You can model temperature difference as voltage drop and heat
transfer as current if you like. Any good engineering heat transfer text
will show you exactly how to do this{but in this case, it won't really
matter all that much}].
--
bz 73 de N5BZ k
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.
bz+ser@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap