1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find amount of H20 needed to cool Steel

  1. Jul 31, 2013 #1
    Hello people, thanks for reading. I am glad you are checking out my problem. Basically I need to confirm a "theory/formula" that my boss had written down from a decade ago declaring how much water (roughly) would be needed to cool Slag from steel making process to below boiling.

    PROBLEM: Figure out how much H20 is needed to cool Slag down to 212°C
    MASS: 2000 lbs. or 1 ton or 907.185 kg
    Carbon Steel: 0.12 (Kcal/Kg°)C
    Heat of H20 Vaporization (Enthalpy): 2260 J/g°C
    T1: 2200°F or 1104.4°C
    T2: 212°F or 100°C
    ΔT: 1104.44 °C
    1 Kcal = 4.184 Kilajoules
    1 Gallon = 3.785 Kg

    (MASS) (HEAT CAPACITY) (ΔTEMPERATURE) = Energy Removed From Slag
    (907.168 Kg) (0.12 Kcal/Kg°)C (1104.44°)C
    (907.168 Kg) (0.12 Kcal/Kg°)C (1104.44°)C
    (120229.5151 Kcal)

    (120229.5151 Kcal) 4.184 Kj
    1 1 Kcal

    503,040.2912 Kj = Energy Needed to be Removed from Slag
    503,040.2912 Kj 503,040,291 Joules

    503,040,291 J 1
    1 2.260 J/g°C

    (222584199.6 Grams) in H20 222,584 Kg

    222,584 Kg 1 gallon
    1 3.785 Kg

    58806.86 Gallons of Water

    Does anyone think I am close to the right solution? Yet again this need not be an extremly precise calculation. I really just want a ballpark that has been found using good math
  2. jcsd
  3. Jul 31, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Check your ΔT calculation for the slag again.
  4. Jul 31, 2013 #3
    Got it, I just had the wrong number written down for the conversion of 2200F to C. I believe my ΔT is ok because I just had a recording error there. Thanks!
  5. Jul 31, 2013 #4
    ΔT is fine, the error is T1, which should have been 1204 C.
  6. Jul 31, 2013 #5
    I think the calculation is fine. What I would normally complain about is that your data in converted units retains absurdly many "significant" digits, with none of them most likely correct; but since you need a ballpark estimate, then you should be fine.
  7. Jul 31, 2013 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    T1 = 2200 F should be 1204 C

    Once you get the energy to be removed (503000 kJ) you use 2.26 J/g-C for the heat of vaporization of water. The correct figure is 2270 kJ/kg-C, which puts your amount of water off by a factor of 1000.
  8. Jul 31, 2013 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Basically, you need to evaporate one 55-gallon drum of water to cool each ton of slag.
  9. Jul 31, 2013 #8
    Indeed. 2260 J/g°C was correct. How did that become 2.260 J/g°C?
  10. Jul 31, 2013 #9
    Thanks everyone! It means a lot to have people helping out. I see where I went wrong. I got ahead of myself and tried to conver 2260 J/g°C (for reasons I dont even remember). I am honestly not a very strong in physics or basic math. I have not taken a course in 5 years and do not keep up enough... I am far removed from even basic mathmatical equations but I am seeing with my new job, I better start remembering. Thanks again everyone. It means a lot
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook