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March 27, 2017

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The specific heat of mercury is .03cal/g*C, and it's boiling point is 357*C. The specific heat of water is 1 cal/g*C. It takes 65 calories of energy to vaporize one gram of mercury and 540 calories to vaporize one gram of water. If both substances begin at about 22*C does it take more energy to boil a gram of mercury or a gram of water? If someone can give me the formula to use, I can try to figure it out myself! Thanks

  • physics - ,

    heat to boil 1 gram water:
    mcdeltaT+ Lv*m
    1*1*(100-22)+540*1

    heat to boil 1 gram Hg
    mcdeltatT+Lv*m
    1*.03(357-22)+65*1

    do the math.

  • physics - ,

    Energy required per gram =
    Q = (heat of vaporization) + (Tboiling - Tinitial)*(specific heat)

    Compute and compare that quantity for both liquids.

    For mercury,

    Q = 65 + 335*0.03

    For water

    Q = 540 + 78*1.0

    Water's heat requirement is much higher

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