A 41 g Ice cube at -21 C is dropped into a container of water at 0 C. How much water freezes onto the ice? The specific heat of ice is .5 ca

Question

A 41 g Ice cube at -21 C is dropped into a container of water at 0 C. How much water freezes onto the ice? The specific heat of ice is .5 cal/g C and it’s heat of fusion of is 80 cal/g.

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Maris 4 years 2021-08-22T05:03:01+00:00 1 Answers 12 views 0

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  1. Tryphena
    0
    2021-08-22T05:04:22+00:00
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    Answer:

    The right solution is “5.38 grams“.

    Explanation:

    The given values are:

    Heat of fusion,

    L = 80 cal/g

    Mass of ice cube,

    m_{ice} = 41 \ g

    Specific heat of ice,

    C_{ice}=0.5 \ cal/g

    Let,

    Gram of water freezes will be “m”.

    ⇒  mL=m_{ice} C_{ice} (0+21)

    Or,

    ⇒     m=\frac{m_{ice} C_{ice} (0+21)}{L}

    On substituting the values, we get

    ⇒         =\frac{41\times 0.5\times 21}{80}

    ⇒         =\frac{430.5}{80}

    ⇒         =5.38 \ grams

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