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27 October, 22:52

How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C?

The specific heat of ice is 2090 J/kg ∙ K, the latent heat of fusion of water is 33.5 * 10^4 J/kg, and the specific heat of water is 4186 J/kg ∙ K

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  1. 27 October, 23:21
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    229,098.96 J

    Explanation:

    mass of water (m) = 456 g = 0.456 kg

    initial temperature (T) = 25 degrees

    final temperature (t) = - 10 degrees

    specific heat of ice = 2090 J/kg

    latent heat of fusion = 33.5 x 10^ (4) J/kg

    specific heat of water = 4186 J/kg

    for the water to be converted to ice it must undergo three stages:

    the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp

    Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J

    the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp

    Q = 0.456 x 33.5 x 10^ (4) = 152760 J

    the water must cool further to - 10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp

    Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J

    The quantity of heat removed from all three stages would be added to get the total heat removed.

    Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J
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