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Homework Help: Reaction Heats

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data

    An aluminum engine block has a mass of 110 kg. If only 20% of the heat produced in the engine is available to heat the block, what mass of octane is required to raise the temperature of this block from 15 celcius to 85 degrees celcius? c(Al) = 0.9 kJ/(kg . degrees celcius)

    2. Relevant equations

    [tex]\Delta[/tex]H=mc[tex]\Delta[/tex]T

    3. The attempt at a solution
    [tex]\Delta[/tex]H=(110kg)(0.9)(85-15)=(110kg)(0.9)(70)=6930 kJ x 0.2 % = 1386 kJ

    Im clueless as to find the mass of octane after this step :(
     
  2. jcsd
  3. Dec 7, 2008 #2

    Borek

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    You will need specific enthalpy for the reaction of octane burning to solve the question.
     
  4. Dec 7, 2008 #3
    The heat of reaction for the combustion of octane is -5517 kJ
     
  5. Dec 7, 2008 #4
    Nope. It's -5517 kJ/mol ... that "per mole" part is important. Presumably if you knew how much heat was produced, you could figure how how many moles of octane -- and if you knew that, from the molecular weight, how many grams of octane.
     
  6. Dec 7, 2008 #5
    okay but none of them give me my answer: THe answer to this question is 0.72 kg

    But anyways here's what ive tried to do

    because only 20% of the heat produced is used to heat the block:

    5517 kJ/mol x (1/1386 kJ) THe kJ part cancels out which gives me: 3.9805 mol and multiply that with 114 g/mol which gives me 453.77 g which is definently not the answer !
     
  7. Dec 7, 2008 #6

    Borek

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    Apart from the fact that it is either 0.2 or 20%, but not 0.2%, are you sure have calculated here total amount of heat that have to be produced from the octane burning?
     
  8. Dec 7, 2008 #7
    The entire Question goes like this:
    Aviation gasoline is almost pure octane [tex]C_{8}H_{18}[/tex]. Octane burns according to the equation:

    [tex]\underbrace{C_{8}H_{18(l)}}_{a} + \underbrace{O_{2(g)}}_{b} \rightarrow \underbrace{CO_{2(g)}}_{c} + \underbrace{H_{2}O_{(l)}}_{d}[/tex]


    [tex]\Delta[/tex][tex]H^{\circ}_{f}[/tex] = -209 kJ for [tex]C_{8}H_{18}[/tex],

    [tex]\Delta[/tex][tex]H^{\circ}_{f}[/tex] = -286 kJ for [tex]H_{2}O[/tex],

    [tex]\Delta[/tex][tex]H^{\circ}_{f}[/tex] = -394 kJ for [tex]CO_{2}[/tex],

    i) Balance the equation
    [tex]C_{8}H_{18}[/tex] + [tex]\frac{25}{2}O_{2}[/tex] [tex]\rightarrow[/tex] [tex]8CO_{2}[/tex] +[tex]9H_{2}O[/tex]

    ii)What is the heat of formation of reactant (b)
    0 kJ because its an element

    iii) Calculate the heat produced when 1.00 L of [tex]C_{8}H_{18}[/tex] is burned. The density of [tex]C_{8}H_{18}[/tex] is 703 g/L.

    (-1) x 1. [tex]8C_{(s)} + + 9H_{2(g)} \rightarrow C_{8}H_{18(l)}[/tex] [tex]\Delta[/tex][tex]H^{\circ}_{f} = -209 kJ[/tex]

    (9) x 2. [tex]H_{2(g)} + \frac{1}{2}O_{2} \rightarrow H_{2}O_{(l)}[/tex] [tex]\Delta[/tex][tex]H^{\circ}_{f} = -286 kJ[/tex]

    (8) x 3. [tex]C_{(l)} + O_{2(g)} \rightarrow CO_{2(g)}[/tex] [tex]\Delta[/tex][tex]H^{\circ}_{f} = -394 kJ[/tex]


    1. [tex]C_{8}H_{18(l)} \rightarrow 8C_{(s)} + 9H_{2(g)}[/tex] [tex]\Delta[/tex][tex]H^{\circ}_{f} = 209 kJ[/tex]

    2. [tex]9H_{2(g)}) + \frac{9}{2}O_{2} \rightarrow 9H_{2}O_{(l)}[/tex] [tex]\Delta[/tex][tex]H^{\circ}_{f} = -2574 kJ[/tex]

    3. [tex]8C_{(l)} + 8O_{2(g)} \rightarrow 8CO_{2(g)}[/tex] [tex]\Delta[/tex][tex]H^{\circ}_{f} = -3152 kJ[/tex]
    ===========================================
    [tex]C_{8}H_{18(l)} + \frac{25}{2}O_{2} \rightarrow 8CO_{2(g)} + 9H_{2}O_{(l)}[/tex] [tex]\Delta[/tex][tex]H^{\circ}_{c} = -5517 kJ[/tex]

    iv)Calculate the heat produced when 1.00 L of [tex]C_{8}H_{18(l)}[/tex] is burned. The density of [tex]C_{8}H_{18(l)}[/tex] is 703 g/L.
    [tex]D = \frac{m}{V}[/tex]

    [tex]703g/L = \frac{m}{1 L}[/tex]

    [tex]m = 703 g[/tex]

    [tex]\frac{703g}{114g/mol} = 6.17 mol[/tex]

    [tex]\frac{5.52 x 10^{3} kJ}{mol}[/tex] x 6.17 mol = 3.4x[tex]10^{4}[/tex] kJ

    v)An aluminum engine block has a mass of 110 kg. If only 20% of the heat produced in the engine is available to heat the block, what mass of octane is required to raise the temperature of this block from 15[tex]\circ[/tex]C to 85[tex]\circ[/tex]C degrees celcius? [tex]c_{Al}[/tex] = 0.9 kJ/(kg [tex]\circ[/tex]C)

    m=110 kg
    c=0.9
    [tex]\Delta[/tex]T=70 (85-15)
    [tex]\Delta[/tex]H=(110)(70)(0.9)=6930 kJ (Total heat released in the aluminum block engine)
     
    Last edited: Dec 7, 2008
  9. Dec 7, 2008 #8
    I dont understand what to do after part v)
     
  10. Dec 21, 2008 #9
    Can somebody please help me out on this ? Ive done all i can
     
  11. Dec 21, 2008 #10
    An aluminum engine block has a mass of 110 kg. If only 20% of the heat produced in the engine is available to heat the block, what mass of octane(C8H18) is required to raise the temperature of the block from 15 degrees celcius to 85 degrees celcius? Specific heat capacity of aluminum = 0.9 kJ/(kg C)

    The enthalpy of combustion of octane is -5517 kJ (the answer is 0.72 kg) But i dont understand how X_X
     
  12. Dec 21, 2008 #11
    Does anyone even care about answering my question ?
     
  13. Dec 22, 2008 #12

    Ivan Seeking

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    Everyone here donates their time. You will have to be patient.
     
  14. Dec 22, 2008 #13

    chemisttree

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    Do it in pieces. How much energy will you need to raise the block by 1 degree C? Then multiply that by delta T, then account for the 20% efficiency.

    You have the skills you need to answer this. Trust what you know and relax.
     
  15. Dec 22, 2008 #14

    Borek

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    That was already done :smile:

    And that's where the problems started...

    ghostanime2001: as I already hinted at, amount of energy needed to heat the block if 20% is used for heating is NOT 6930 kJ x 0.2.
     
  16. Dec 22, 2008 #15
    Then what do i do....... ill say it again ive done ALL i can.... i have no more ideas.
     
  17. Dec 22, 2008 #16

    chemisttree

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    Borek, humor me.:smile:


    Ghost, so tell me, how much energy do you need to raise the engine block by one degree C?
     
  18. Dec 22, 2008 #17
    deltaH=(110)(70)(1) = 99 kJ
     
    Last edited: Dec 22, 2008
  19. Dec 22, 2008 #18

    chemisttree

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    [tex]\Delta[/tex]H=mc[tex]\Delta[/tex]T

    [tex]\Delta[/tex]T = 1 (if you want the energy per degree C)... you need 'c'
     
  20. Dec 22, 2008 #19
    so i am right. u said the energy per degree ive given you the energy per degree man.. what else do u want yo WHy in the world would i need C man ????????? i dont need the heat capacity of octane dawg
     
    Last edited: Dec 22, 2008
  21. Dec 22, 2008 #20

    chemisttree

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    Uhhhh, I guess. I just thought that you were trying to do too much at once and when you made a mistake it looked too complicated for you to figger out.... dawg.

    You should have written:

    [tex]\Delta[/tex]H=(110 kg)(0.9 kJ/kg degrees C)(1 C)= 99 kJ (heat required to raise block by 1 C)
     
  22. Dec 22, 2008 #21
    okay fine... now what
     
  23. Dec 22, 2008 #22
    is this a calorimetry kinda question ? the aluminum engine block is the calorimeter and the octane in the substance acting like water ?
     
  24. Dec 22, 2008 #23

    chemisttree

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    No, it isn't calorimetry.

    Now you need to find the energy required to raise it 70 C. I think you know this already, so I'll tell you what comes after that. When you have that answer, you need to think of how much energy must be put into the engine at 20% efficiency to equal the energy required to raise the temperature by 70C. Like Borek said, that's where you made a minor error before...

    Remember, you will need to add more energy than you need to raise the block by 70C if only 20% of it goes into heating the block.

    You are going to kick yourself when you see how easy it is.
     
  25. Dec 22, 2008 #24
    just tell me how to do it im sick and ****ing tired of this question already
     
  26. Dec 22, 2008 #25

    chemisttree

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    Ghost, you are obviously skilled, so you can do this. I'm trying to show you how to break down a problem into pieces when you are at your wits end, so bear with me.

    You only made a minor mistake. You have to find it now.
     
    Last edited: Dec 22, 2008
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