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14 May, 07:20

A 2.9 kg lump of aluminum is heated to 92 C and then dropped into 11.0 kg of water at 5.2 C. Assuming that the lump-water system is thermally isolated, what is the system's equilibrium temperature? Assume the specific heats of water and aluminum are 4186 and 900

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  1. 14 May, 07:30
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    9.85 C

    Explanation:

    Let T be the temperature off equilibrium mixture.

    Heat loss or gain = mass x specific heat x change in temperature

    Heat loss by lump of Aluminium = 2.9 x 900 x (92 - T)

    Heat gain by water = 11 x 4186 x (T-5.2)

    Heat loss = heat gain

    2.9 x 900 x (92 - T) = 11 x 4186 x (T - 5.2)

    T = 9.85 C
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