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4 July, 13:13

330 grams of boiling water (temperature 100°C, specific heat capacity 4.2 J/K/gram) are poured into an aluminum pan whose mass is 840 grams and initial temperature 29°C (the specific heat capacity of aluminum is 0.9 J/K/gram). After a short time, what is the temperature of the water?

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  1. 4 July, 15:53
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    T = 74°C

    Explanation:

    Given Mw = mass of water = 330g, Ma = mass of aluminium = 840g

    Cw = 4.2gJ/g°C = specific heat capacity of water and Ca = 0.9J/g°C = specific heat capacity of aluminium

    Initial temperature of water = 100°C.

    Initial temperature of aluminium = 29°C

    When the boiling water is poured into the aluminum pan, heat is exchanged and after a short time the water and aluminum pan both come to thermal equilibrium at a common temperature T.

    Heat lost by water equal to the heat gained by aluminium pan.

    Mw * Cw * (100 - T) = Ma * Ca * (T-29)

    330*4.2 * (100 - T) = 890*0.9 * (T-29)

    1386 (100 - T) = 801 (T - 29)

    1386/801 (100 - T) = T - 29

    1.73 (100 - T) = T - 29

    173 - 1.73T = T - 29

    173+29 = T + 1.73T

    202 = 2.73T

    T = 202/2.73

    T = 74°C
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