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Heating water with carbon gas

  1. Oct 10, 2014 #1
    1. The problem statement, all variables and given/known data
    200 liters of water are heated from 100 to 650C by burning coal gas. The heat loss in the chimney is 25%.
    How many cubic meters of gas are needed

    2. Relevant equations
    Calories required: [itex]Q=mc\delta t[/itex]
    Burning heat of coal gas: 4320[Cal/m3]

    3. The attempt at a solution
    Net heat needed: [itex]Q=200\cdot 55=11,000[/itex]
    Gross heat needed: [itex]Q\cdot 0.75=11,000\rightarrow Q=14,667[/itex]
    Gas needed: [itex]4320\cdot x=14,667\rightarrow x=3.4[m^3][/itex]
    It should be 2.9
     
  2. jcsd
  3. Oct 10, 2014 #2

    SteamKing

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    You need to show units in all your calculations. It makes your work much easier to follow and analyze for mistakes.
     
  4. Oct 10, 2014 #3
    Net heat needed: [itex]Q=200[kg]\cdot 55^0=11,000[Cal-kg][/itex]
    Gross heat needed: [itex]Q[Cal-kg]\cdot 0.75=11,000[Cal-kg]\rightarrow Q=14,667[Cal-kg][/itex]
    Gas needed: [itex]4320[Cal/m^3]\cdot x=14,667[Cal-kg]\rightarrow x=3.4[m^3][/itex]
    It should be 2.9[/QUOTE]
     
  5. Oct 10, 2014 #4
    You are missing something in the first formula. How can you get Calories by multiplying kg by degrees?
     
  6. Oct 10, 2014 #5

    mfb

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    And how did the kg vanish again in the third line? This issue will go away once you fix the first line.

    What should be 2.9?
     
  7. Oct 10, 2014 #6
    I mistakenly omitted the specific heat:
    Net heat needed: [itex]Q=200[kg]\cdot 1[\frac{Cal}{^0\cdot kg}]\cdot 55^0=11,000[Cal][/itex]
    Gross heat needed: [itex]Q[Cal]\cdot 0.75=11,000[Cal]\rightarrow Q=14,667[Cal][/itex]
    Gas needed: [itex]4320[Cal/m^3]\cdot x[m^3]=14,667[Cal]\rightarrow x=3.4[m^3][/itex]
    The answer should be 2.9[m3]
     
  8. Oct 11, 2014 #7

    NTW

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    Your answer is right, IMHO. Let's solve the problem by a false-position approach: we assume that you burn 2 cu. m of gas. The energy obtained is 4320*2 = 8640 Cal. You lose 25% through the chimney, and are left with 8640*0,75 = 6480 Cal. Now, you have 200 liters of water, so with those 6480 Cal, its temperature will rise by 6480/200 = 32,40ºC. But you need a rise of 55ºC, hence you need more gas... Exactly (55/32,4)*2 = 3,4 cu. m...
     
  9. Oct 11, 2014 #8
    Thanks
     
  10. Oct 11, 2014 #9

    mfb

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    ° is not a unit. Use K instead.
    Apart from that, it looks fine (but I really don't like "Cal" for kcal).
     
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