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13 November, 05:37

Steam (water vapor) at 100 degrees Celsius is added to a thermally insulated container with 200 kg of ice at zero degrees Celsius. The final mixture is water at 30 degrees Celsius. What was the initial mass of the steam?

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  1. 13 November, 05:54
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    m = 359.24 kg

    Explanation:

    given dа ta:

    mass of ice = 200 kg

    latent heat of steam = 2260 kJ/kg

    latent heat of ice = 334 kJ/kg

    from conservatioon of energy principle we know that

    heat lost by steam will be equal to heat gained by ice

    therefore we have

    (ml + mcdt) = (ml + mcdt)

    m*2.26*10^6 + m*4186 * (100-30) = 200*3.33 * 10^6 + 2000*4186 * 30

    m = 359.24 kg
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