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11 February, 13:57

Your 300 mL cup of coffee is too hot to drink when served at 87.0 degrees Celsius. What is the mass of the ice cube, taken from a - 15.0 degrees Celsius freezer, that will cool your coffee to a pleasant 65 degrees Celsius?

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  1. 11 February, 16:19
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    m=43.39 Kg

    Explanation:

    from conservation principle we have

    heat gained by ice equal to heat lost by coffee

    heat gained by ice = heat from - 15 °C ice to 0 °C ice + 0 °C ice to 0 °C water + 0 °C water to 65 °C water

    = m*S1*15 + mL + ms2 * 65

    where s is specific heat of respective phase

    = m (2.08 * 15 + 334.16 + 4.184 * 65)

    = m*637.32

    heat lost by the coffee = ms * delta T

    =0.3 * 4190 * (87-65)

    =27654

    therefore

    27654 = m*637.32

    m=43.39 Kg
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