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How many calories are required to evaporate 1 gram of boiling water?

Also, how many calories of heat are lost by water freezing at 0 degrees celsius?

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One Response to “How many calories are required to evaporate 1 gram of boiling water?”

  1. BP said:

    This is a two step process.
    1) heat the water to boiling point
    2) phase change from liquid to gas

    Q= cm(dtT)
    Q= mL, where L is the latent heat
    c-water= 4.184 j/g
    c-vapor = 2258 j/g

    1) Q = 4.184j/g * 1g*100degrees
    2) Q = 1*2258
    418.4j+2258j = 2,676.4j or 639.2 cal

    1j=4.18 cal




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