To treat a burn on your hand, you decide to place an ice cube on the burned skin. The mass of the ice cube is 16.0 g, and its initial temperature is -11.4 °C. The water resulting from the melted ice reaches the temperature of your skin, 31.0 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand.

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  1. Do you have the heat of fusion? The specific heat of ice? The specific heat of H2O?

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  2. Enthalpy of fusion
    333.6 J/g
    6010. J/mol

    Specific heat of solid H2O (ice)
    2.087 J/(g·°C)
    37.60 J/(mol·°C)

    Specific heat of liquid H2O (water)
    4.184 J/(g·°C)
    75.37 J/(mol·°C)

    Specific heat of gaseous H2O (steam)
    2.000 J/(g·°C)
    36.03 J/(mol·°C)

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  3. Heat absorbed by ice raising T from -11.4 to zero.
    q1 = mass ice x specific heat ice x (Tf-Ti).
    q1 = 16.0g x 2.087 J/g x 11.4 = ?J

    q2 = heat to melt the ice to liquid at zero C.
    q2 = mass x heat fusion ice.
    q2 = 16.0g x 333.6 J/g = ?

    q3 = heat to raise T of melted ice at zero C to 31.0 C.
    q3 = mass H2O x specific heat H2O x (Tf-Ti)
    q3 = 16.0g x 4.184 J/g*C x 31 =?

    Total heat absorbed = q1 + q2 + q3 = ?

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