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A 21.0 g sample of aluminum, which has a specific heat capacity of 0.897 J g °C , is dropped into an insulated container co

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Answer #1

heat lost by aluminium = heat gained by water

qlost = qgain

mCAl\DeltaT = mCwater\DeltaT

21.0 x 0.897 x (90.1 -X) = 200 x 4.186 x (X - 25.0)

18.837 (90.1 -X) = 837.2 (X - 25.0)

1697.21 - 18.837 X = 837.2 X - 20930

22627.21 = 856.04 X

X = 26.40C

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