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## btu math help

If 11532 btu's were removed from 62 lbs of water at a 72 degrees f water temp , what would be the new temp of the water ?
Homework Question....appreciate some help with a formula.
Thanks

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It takes 1 BTU to raise 1 pound of water 1 degree F.

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understand the definition, still confused on how to answer the question.
Thanks for the response

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Either there is a typo here or you're getting into phase change, too.

You're removing eleven thousand, five hundred and thirty two BTUs from sixty two pounds of water at seventy two degrees?

I typed the full words out to prevent confusion.

I ran a little math and this doesn't add up to a typical problem.

Are you adding or removing heat?

Also, moving thread to tech to tech. AOP is for equipment questions.

5. Originally Posted by oliesteveo
If 11532 btu's were removed from 62 lbs of water at a 72 degrees f water temp , what would be the new temp of the water ?
Homework Question....appreciate some help with a formula.
Thanks
11532/62=186 per pound. 40 gets to freezing.

How many btu to freeze a pound? probably more than 146, so I'd say new water temp 32f.

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yes sir...removing 11532 btu from 62 lbs of 72 degree f water there is a change of state latent and sensible i think
here is another one 13080 btu were removed from 10 lbs of water at 218 degree f what would be the temp of the water

Thanks for the help

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Originally Posted by oliesteveo
yes sir...removing 11532 btu from 62 lbs of 72 degree f water there is a change of state latent and sensible i think
here is another one 13080 btu were removed from 10 lbs of water at 218 degree f what would be the temp of the water

Thanks for the help
water at 218???

or STEAM at 218?

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Originally Posted by tedkidd
11532/62=186 per pound. 40 gets to freezing.

How many btu to freeze a pound? probably more than 146, so I'd say new water temp 32f.
This is it. You'll wind up with a water/ice slurry at 32 degrees.

it will take 7440 BTU to drop the 62# from 72 to 32 (186 * 40) = 7440 This leaves 4092 extra BTUs for latent conversion

You'll wind up with about 28# of ice by the time it's all done.

9. Originally Posted by oliesteveo
If 11532 btu's were removed from 62 lbs of water at a 72 degrees f water temp , what would be the new temp of the water ?
Homework Question....appreciate some help with a formula.
Thanks
Formula: BTU's = Lbs x Deg x SH (specific heat of material)

Note: specific heat of water is 1.0.

So with a little Algerbra: Btu χ (Lbs x SH) = Deg.

Now to calculate Btu's required to lower temp from 72Ίf to 32Ίf (ICE), or 40Ίf.

62 lbs x 40 deg x 1 SH = 2480 btu's

11532 - 2408 = 11408 btu's left at 32Ίf

Change of State - Latent heat of Fusion

144 btu/lb to change liquid to ice : 62 lbs x 144 btu's = 8928 btus

Now we have used : 2480 (lowering temp to 32Ίf) + 8928 (change of state) = 11408 btus

11532 - 11408 = 124 btus left

Note: specific heat of ice is 0.5

124 Btu's χ (62 Lbs x 0.5 SH) = 4 degs

32Ίf (ice) - 4Ίf = 28Ίf (final temperature)

10. I converted to per pound because it was easier for me to follow, would have converted back at the end had I known its only 144 to ice. Guess I assumed more because there is so much energy in condensation.

BTW, What is that exact number?

Did not realize its 1 btu to drop ice 2 f, so would said 30 and gotten the answer wrong even knowing cost of getting to ice. Thanks for the formula and explanation on specific heat!

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change of state at 212 degrees f 970 btu at this point

12. I remember in physics we had a question:
A round tub with 2' height, made of standard steel has a volume of 50 gallons at 50*, and it is filled with 15 gallons of 50*F water. How many gallons of boiling water would it take to over flow the tub, and at what temp would the water be? Answer to the nearest 0.001gal and 0.1*F.

13. I'm sure it's not as simple as 35.001 gallons and 98.6, and am dying to see the formula...

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