Tilted Forum Project Discussion Community - View Single Post - [Math Problem] 80 degree hot tub.. how much boiling water to make it 95+ degrees?

Yakk · 06-19-2004, 08:30 AM

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.

06-19-2004, 08:30 AM	#4 (permalink)
Yakk Wehret Den Anfängen! Location: Ontario, Canada	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: 500T - bT + 100b --------------------------- 500 Simplify more: T + 1/5b - bT/500 or, the change in tempurature is: b/5 - bT/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) + b212] / 500 T_after = T_before + b212/500 - T_beforeb/500 T_change = b212/500 - T_beforeb/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. __________________ Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.