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Need help on solving heat problems

  1. Mar 19, 2006 #1
    I'm a high school student and I have a test soon. I know the basic equation we are using (Q=mCdeltaT), but I having extreme difficultly employing them to get the answer I need, I'm currently stuck on this practice problem:


    A 0.925 KG mass of water is place in a container (mass .156 KG) with a specific heat of 903 J/Kg C, The initial temperature is 26.0 C. A .255 KG piece of metal (95.0C) is dropped into the water. The final temperature of the water is 31.5 C.

    What is the specific heat of the metal?


    Help, please?

    I have more problems, and if you have the time, it would be nice if you could AIM me at "CannonballSU".
     
  2. jcsd
  3. Mar 19, 2006 #2
    Second question:

    A container holds 525 grams of water at 100C
    How much heat energy must be gained by the water to change it into 100C steam?


    Is it 0 Joules due to the fact that it is already at 100C?
     
  4. Mar 19, 2006 #3
    Show some work! :smile:
     
  5. Mar 19, 2006 #4
    Okay. I figured this out, actually, it takes 1,186,500 Joules (Q = .525KG (2260000 J) (disregarding sig figs), because the added thermal energy for a phase change does not increase the kinetic energy of the particles...right?
     
  6. Mar 19, 2006 #5
    explanation

    A 0.925 KG mass of water is place in a container (mass .156 KG) with a specific heat of 903 J/Kg C, The initial temperature is 26.0 C. A .255 KG piece of metal (95.0C) is dropped into the water. The final temperature of the water is 31.5 C.

    What is the specific heat of the metal?

    You are right, the formula is as follows:

    [tex]\triangle Q = m C_p \triangle T[/tex]

    Where Q is heat, m is mass in kg, C(p) is Specific Heat and delta T is change in temperature in Celsius. In this situation, I'm ignoring the container as it seems to be stupid.. Never had to solve one that included mass of container :-|

    let us note that temperature always returns to equilibrim. In this case, the final temperature of the water is equal to the final temperature of the metal.

    So the equation is as follows where (W) is water and (M) is metal (parantheses means subscript)

    [tex]\triangle Q_W = \triangle Q_M [/tex]
    [tex]m_W C_W \triangle T_W = m_M C_M \triangle T_M[/tex]
    [tex]m_W C_W (T_f - T_i) = m_M C_M (T_f - T_i)[/tex]

    The final temperature is the same in both situations, we plug in:

    [tex](.925) (4.184) (31.5- 26.0) = (.255) C_M (31.5 - 95.0)[/tex]

    Solve for C(m) o_O Note you may get a different answer if you actually do include the mass of the container...
     
  7. Mar 19, 2006 #6
    Slow down. Explain what you are doing, and show your steps. I have no clue how you are jumping to your answers.

    Please do NOT give complete solutions in the future Da-Force.

    Also, Please stop giving WRONG solutions. Your above statement is wrong.
     
    Last edited: Mar 19, 2006
  8. Mar 19, 2006 #7
    It's not the complete solution as I never seen one that included the mass of the container.. I've explained my reasoning before equations.
     
  9. Mar 19, 2006 #8
    Okay, this is what I did

    I first found the heat of the water

    Q = .925(4180)(31.5-26.0)
    Q = 21265.75

    Then I plugged Q into the equation for the metal:

    21265.75 = .255 C (31.5-95)

    Which got me C = 1313 J/kg C

    How do I take in account of the container's specific heat? (I'll be back after grabbing dinner)
     
  10. Mar 19, 2006 #9
    Before we even do this problem with any numbers, lets step back and talk about what is going on in this process. Please explain to me in your words EXACTLY what happens. You will then find the solution easy.
     
  11. Mar 19, 2006 #10
    Since I know that the amount of energy within the system remains the same. I can apply my value of 21265.75 joules to any equation dealing with the cup, metal or water. So this means, when I place the brick into the water, which is held in the cup, the energy, due to entropy, is being evenly distributed within the system, which makes the hotter objects (the metal) colder, and the colder, warmer.

    So...the amount of energy the cup absorbes doesn't really affect my caculation for the specfic heat of the metal, does it?
     
  12. Mar 19, 2006 #11
    I thought you just said hot objects get cold and vice versa? Is the metal container some special exception?
     
  13. Mar 19, 2006 #12
    Okay, I get this.

    I caculate the Q for the container and subtract that from the heat of water and take that answer and plug it in to find the specific heat of the metal. Right?
     
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