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29 August, 02:16

Two 20.0 g ice cubes at - 20.0 ∘ C are placed into 285 g of water at 25.0 ∘ C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, T f, of the water after all the ice melts.

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  1. 29 August, 03:56
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    Ft = 17.48°C

    Explanation:

    Ft is the final temperature. However, ice absorbs heat during two process of melting and cooling and as such, there is no loss of heat to or from the surrounding hence by conservation of energy.

    Therefore,

    Heat absorbed by water of 20g = heat rejected by water of 265g.

    So; M (ice) [C (ice) [ (ΔT) + LH (ice) + C (water) (ΔT) ] = C (water) M (water) (ΔT)

    So, 20[ (2.108) [0 - (-20) ] + 333.5 + 4.187 (Ft - 0) ]] = (285) (4.187) (25 - Ft)

    To get;

    7513 + 83.74 Ft = 29832.4 - 1193.3 Ft

    So factorizing, we get;

    83.74 Ft + 1193.3 Ft = 29832.4 - 7513

    So; 1277.04 Ft = 22319.4

    So; Ft = 22319.4/1277.04 = 17.48°C
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