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14 June, 02:23

How to solve determine the final temperature when 450.2 grams of aluminum at 95.2°c is placed in an insulated calorimeter with 60.0 grams of water at 10.0°c?

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  1. 14 June, 02:46
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    To solve this, we simply equate the change in enthalpy for the two substances since heat gained by water is equal to heat lost of aluminum. We know that the heat capacity of aluminum is 0.089 J/g°C and that of water is 4.184 J/g°C. Therefore:

    450.2 (95.2 - T) (0.089) = 60 (T - 10) (4.184)

    3,814.45456 - 40.0678 T = 251.04 T - 2,510.4

    291.1078 T = 6,324.85456

    T = 21.7°C
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