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27 May, 19:32

How much heat is required to vaporize 33.8 g of water at 100 degrees Celsius

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  1. 27 May, 19:47
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    Answer : Heat required will be 76.42 KJ

    Explanation : Here we have the data as 33.8 g of water and temperature is 100°C

    To calculate the heat required we need to use the formula as,

    q = mol of water / heat of vaporization of water.

    So we have convert the mass into moles,

    33.8g / 18g = 1.877 moles of water.

    and heat of vaporization is 40.7 KJ

    Now, q = 1.877 X 40.7 KJ = 76.4 KJ

    So the heat required will be 76.4 KJ
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