View RSS Feed

The 5.5 Gallon Nano tank

Basic college chemistry? Anyone??

Rate this Entry
6 Comments
by
Douwant2play
, 05-04-2010 at 11:23 AM (3734 Views)
So I have a chemistry final today and I am stuck on one of my review problems and the prof. doesn't want to help me set this up to get the correct answer, I believe because a similar problem will be on the exam. Anyway if anybody is a chem major/teacher/enthusiast/just knows some chemistry can you help me out? Here is the problem that has kept me awake the past few nights (and for once its not the fault of my tank!):

A 100.g iron ball(specific Heat = 0.473 J/g*C) is heated to 125*C and is placed in a calorimeter holding 200 g of water at 25*C. What will be the highest temperature reached by the water?

(* = Degrees symbol)

thanks in advance for the help

Submit "Basic college chemistry? Anyone??" to Digg Submit "Basic college chemistry? Anyone??" to del.icio.us Submit "Basic college chemistry? Anyone??" to StumbleUpon Submit "Basic college chemistry? Anyone??" to Google

Categories
‎ Random Thoughts

Comments

  1. melev's Avatar
    When you do figure it out, tell us the answer.

    I thought you were going to ask what color spectrum is the iron ball once heated, and then what color will it be in the water.
  2. freewest's Avatar
    I'd say you set up the problem like

    100g (0.473 J/g C) (125 C) + (200g) (4.186 J/g C) (25 C) = 100g (0.473 J/g C) (T) + (200g) (4.186 J/g C) (T)

    then solve for T. I'm an engr major though so see if it makes sense yourself.
  3. Douwant2play's Avatar
    I got it!!!!

    mass of Fe=100g
    CP=Specific heat of Fe= 0.473J/gC
    Temp1 Fe= 125C
    mass of Water= 200g
    Cp=specific heat of Water= 4.184J/gC <---- I knew I was missing something this was it! Thanks Freewest! Such an easy thing to overlook!
    Temp 1 of water= 25.0C
    T2= ?


    (-100g)(-0.473 J/gC)(-T2 + 125C) = (200g)(4.184J/gC)(T2 - 25.0C)

    -47.3J/C T2 + 5913J = 836.8J/C T2 - 20920J

    5913J + 20920J = 836.8J/C T2 + 47.3J/C T2

    26833J = 884.1J/C T2

    26833J/884.1J/C = T2

    30.35C = T2

    *Deep sigh of Relief*

    Thank goodness this is my last Chem. Exam!
  4. freewest's Avatar
    no problem. if your chem teacher is particular about how you set up problems you probably want it set up like this.

    Q Fe = -Q H20
    (100g)(0.473 J/gC)(T2 - 125C) = -(200g)(4.184J/gC)(T2 - 25.0C)

    good luck on the exam
  5. Douwant2play's Avatar
    Thanks everyone! After that exam It's time for wings and beer! Wingstop heeeeeeeeere I come!
  6. Douwant2play's Avatar
    oh and that question was on the exam, actually 3 of them!