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If a piece of aluminum with mass 3.90 g and a temperature of 99.3° C is dropped into 10.0 cm^3 of water at 22.6° C, what will be the final temperature of the system? (Density of water is 1.00 g/cm^3). Also the specific heat (Cp) of aluminum (Al) is .903 J/g°C

2007-05-15 09:03:16 · 1 answers · asked by young wiz 2 in Science & Mathematics Chemistry

1 answers

The heat gained by the water must equal the heat lost by the aluminum. The formula is mCΔT, where m is mass, C is specific heat, and ΔT is change in temperature. Our only unknown is the final temperature, T. We know the specific heat of water to be 4.18 J/g°C, even though it wasn't given in the problem.

The heat gained by the water will be (10 g)(4.18 J/g°C)(T - 22.6). The heat lost by the aluminum will be (3.90 g)(0.903 J/g°C)(99.3 - T). So (10)(4.18)(T - 22.6) = (3.90)(0.903)(99.3 - T), which is now a straightforward algebra problem.

2007-05-15 09:06:17 · answer #1 · answered by DavidK93 7 · 0 1

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