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Heat value problem

  1. Nov 26, 2003 #1
    ok...a burner containing 16.95g liquid fuel is used to heat 200g water in a beaker from 25 degrees to 44 degrees. assuming that half the heat produced is lost to its surroundings, calculate the heat value of the fuel in kg to the negative gram....

    some chemistry problem but i cant find anywhere to do it
  2. jcsd
  3. Nov 26, 2003 #2


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    Re: problem

    LOL that is so damn easy simply because it is worded badly.
    Look at what it is asking for "calculate the heat value of the fuel in kg to the negative gram".
    It is telling me that my units are to be kg/g. Since there is always 1kg/1000g, my answer is simply 1kg/1000g

    However, the question might actually want KJ/g so let's try that.
    First, calculate the energy that the water gained.
    Second, double that energy because only half of the total energy was put into the water.
    Third, divide the total energy by the mass of the fuel in grams.

    Scroll way down to compare your work with mine.

    The total heat put into the water
    E = mcT
    E = (0.2Kg)(4.19KJ/Kg)(44-25)
    E = 15.922kJ in the water

    It says that only half of the heat produced was used in the water. That means
    E = (2)(15.922)
    E = 31.844KJ total

    To get KJ/g, just divide those out.
    31.844KJ/16.95g = 1.8787KJ/g
    = 1.88KJ/g
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