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06-18-2004, 02:44 PM | #1 (permalink) |
Banned from being Banned
Location: Donkey
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[Math Problem] 80 degree hot tub.. how much boiling water to make it 95+ degrees?
No, this isn't a homework question. It's really something I've honestly wondered but couldn't exactly figure out.
There have been times where I forgot to turn my hot tub temperature up (I keep it around 75 to conserve elec.).. and have wondered how many 1 gallon pots of boiling water it would take to raise the temperature 20 degrees. There was a time in the winter when the heater tripped, and due to the design, the heater won't turn on unless the water is at 40 degrees. Well, the water was about 36-38 degrees and I actually did think about putting pots of boiling water.. and how many it would take to raise that temp up to 40. Yes, I'm retarded
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06-18-2004, 07:51 PM | #2 (permalink) |
Psycho
Location: Atlanta, GA
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I do believe that's a question for thermodynamics. However, I think we'd have to know how much water there was at the lower temperature before we could answer the question.
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06-19-2004, 08:30 AM | #4 (permalink) |
Wehret Den Anfängen!
Location: Ontario, Canada
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I think water has a relatively flat heat capacitance?
So, (500-b-z)*T + (100-epsilon)*b -------------------------------------- 500-z should be the tempurature of the water if b is the amount of boiling water z is the amount of empty space left over epsion is the amount the boiling water cooled down before you pored it into the pool. T is the current tempurature of the pool in celcius. We can be lazy, and assume epsilon and z are 0. (500-b)*T + (100)*b -------------------------------------- 500 Simplify: 500*T - b*T + 100*b --------------------------- 500 Simplify more: T + 1/5*b - b*T/500 or, the change in tempurature is: b/5 - b*T/500 Refactored: b* (1/5 - T/500) each gallon of boiling water increases the tempurature by about (1/5-T/500) degrees. If the water is already boiling, this means the water goes up by (1/5 - 100/500) = 0 degrees. If the water is warmish (30 or so), the water goes up 1 degree for every 7 gallons of water you add. If the water is cold (near freezing), the water goes up 1 degree for every 5 gallons of water. Hmm. I'm now thinking you aren't using celcius. =) Lets do it again in silly-units(tm). Water boils at 212. So, T_after = [(T_before)*(500-b) + b*212] / 500 T_after = T_before + b*212/500 - T_before*b/500 T_change = b*212/500 - T_before*b/500 T_change = b * (212/500 - T_before/500) T_change = b * (212-T_before) / 500 where b is the amount of boiling water added, and T_before is the tempurature of the pool before (in degrees F). So, water at 35 and you want it at 40 means 5 = b * (212-35)/500 b = 14 gallons approximetally water is at 75, and you want it at 95: 20 = b * (212-75)/500 b = 73 gallons approximetally.
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06-19-2004, 10:09 AM | #6 (permalink) |
Reclusiarch
Location: Unfortunately Houston, TX
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I've had to solve problems like this in my chemical engineering course before.
It has a lot to do with energies going into and out of the system. I could figure it out, but I'm not ambitious enough.
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06-21-2004, 02:56 PM | #8 (permalink) |
Junkie
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for 500 gallons from 75 to 95, my answer was 85 gallons:
Energy balance: m1*Cp*(Tf - Ti) = m2*Cp(Tb-Tf) m1 = mass of water in tub Cp = heat capacity (assumed constant) Tf = 95F Ti = 75f m2 = mass of water needed Tb = 212 The equation reduces to: m2 = m1*(Tf - Ti)/(Tb - Tf) a cool thing about this eq is that it as long as the volume is not a funtion of temperature (you can assume it is not) you can use any units you want to use for the temp. You can also interchange volume and mass (using any units) and get the same answer. Last edited by kutulu; 06-21-2004 at 03:07 PM.. |
06-23-2004, 07:36 AM | #9 (permalink) | |
Tilted
Location: Starkvegas
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Quote:
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boiling, degree, degrees, hot, make, math, problem, tub, water |
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