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28 December, 13:09

Two 20.0-g ice cubes at - 13.0 °c are placed into 255 g of water at 25.0 °c. assuming no energy is transferred to or from the surroundings, calculate the final temperature, tf, of the water after all the ice melts.

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  1. 28 December, 13:14
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    Let's apply the concept of conservation of energy.

    ∑E = 0

    Energy released by ice + Energy absorbed by water = 0

    Energy released by ice = - Energy absorbed by water

    (mass of alloy) (Cp, ice (Tf - T₀) = - (mass of water) (Cp, water) (Tf - T₀)

    where Cp is the specific heat capacity

    *Cp for ice is 2.108 J/g·°C

    *Cp for water is 4.184 J/g·°C

    Substituting the values,

    (2*20 g) (2.108 J/g·°C) (Tf - - 13) = - (255 g) (4.184 J/g·°C) (Tf - 25)

    Solving for Tf,

    Tf = 22.22 °C
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