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8 May, 10:58

Calculate the heat change in calories for condensation of 20.0 g of steam at 100∘C.

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  1. 8 May, 12:13
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    The latent heat of vaporization of steam is equal to its latent heat of condensation. The heat released is given by:

    Q = ml

    Q = 20 x 2257 joules/gram

    Q = 45.14 kJ of heat is removed from the steam

    1 calorie = 4.184 kJ

    45.14 kJ = 10.8 cal of energy

    The energy removed is 10.8 calories
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