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5 June, 15:30

A 455 g piece of bopper tubing is heated to 89.5 C and placed in an insulated vessel containing 159 g of water at 22.8 C. Assuming no loss of water and a heat capacity of 10. J/K for the vessel, what is the final temperature (c of copper = 0.387 J/g*K) ?

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  1. 5 June, 16:00
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    The final temperature of the setup = 36.6°C

    Explanation:

    Let the final temperature of the setup be T

    Heat lost by the copper tubing = Heat gained by water and the vessel

    Heat lost by the copper tubing = mC ΔT = 455 * 0.387 * (89.5 - T) = (15759.61 - 176.1T) J

    Heat gained by water = mC ΔT = 159 * 4.186 * (T - 22.8) = (665.6T - 15175.1) J

    Heat gained by vessel = c ΔT = 10 * (T - 22.8) = (10T - 228) J

    Heat lost by the copper tubing = Heat gained by water and the vessel

    (15759.61 - 176.1T) = (665.6T - 15175.1) + (10T - 228)

    851.7 T = 31162.71

    T = 36.6°C
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