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11 July, 00:31

If 18.0 kJ of energy are applied to 250. grams of water at 37°C, what will be the final temperature

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  1. 11 July, 04:26
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    54.1°C

    Explanation:

    Now we have the following parameters from the question;

    Heat applied (H) = 18.0*10^3 Joules

    Mass of the water (m) = 250g = 0.25Kg

    Initial temperature of the water (θ1) = 37°C

    Final temperature of the water (θ2) = the unknown

    Heat capacity of water (c) = 4200JKg-1

    From;

    H = mc (θ2-θ1)

    Substituting values appropriately

    18*10^3 = 0.25 * 4200 (θ2-37)

    18*10^3 = 0.25 * (4200θ2 - 155400)

    18*10^3 = 1050θ2 - 38850

    18*10^3 + 38850 = 1050θ2

    56850 = 1050θ2

    θ2 = 56850/1050

    θ2 = 54.1°C
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