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24 February, 11:09

Suppose 45.0 g of water at 85 °C is added to 105.0 g of ice at 0 °C. The molar heat of fusion of water is 6.01 kJ/mol, and the specific heat of water is 4.18 J/g °C. On the basis of these data, (a) what will be the final temperature of the mixture and (b) how many grams of ice will melt?

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  1. 24 February, 12:00
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    A. Final Temp = 36.428C and

    47.9g ice will melt

    Explanation:

    Given the following dа ta:

    Mass of water (M1) = 45.0g = 0.045kg

    Temperature (T1) = 85C = 358k

    Mass of ice (M2) = 105g = 0.105kg

    Temperature (T2) = 0c = 273k

    Specific heat of water (C) = 4.18j/gC = 0.00418kj/kgc

    Molar heat of fusion of water = 6.01kj/mol

    Therefore, heat required (q) = MCT

    M1C (T1-T2) = M2C∆T

    By putting the data we have

    0.045*0.00418 * (358-273) = 0.105*0.00418*∆T

    ∆t = 0.045*0.00418*85/0.105*0.00418

    ∆t = 36.428C

    Gram of ice that would melt would be 47.9g
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