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14 December, 03:16

A 1.00x10^2-g aluminum block at 100.0 degrees celsius is placed in 1.00x10^2g of water at 10.0 degrees celsius. the final temperature of the mixture is 26.0 degrees celsius. what is the specific heat of the aluminum

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  1. 14 December, 03:22
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    Specific heat capacity of a substance is the amount of heat needed to raise the temperature of a mass of 1 kilogram by 1 kelvin. it is calculated by dividing the quantity of heat by the product of mass (kg), change in temperature. When more than one substances are involved where one is gaining heat and the other is losing heat then we equate the heat lost to the heat gained.

    That is; Heat gained = heat lost

    In this case, Heat gained by water = heat lost by aluminium.

    Quantity of heat is given by the product of specific heat capacity, mass and change in temperature, thus H = m*c*Ф, where H is heat, c is the specific heat capacity and Ф temperature change.

    Heat lost by aluminium = 0.1 kg * c (unknown) * 74 degrees (temperature change). We get 7.4 c joules.

    Heat gained by water = 0.1 kg * 4200 J/kg/k (specific heat capacity of water) * 16 degrees (temperature change). we get 6720 joules.

    Therefore, 7.4 c = 6720

    c = 908.108 J/kg/k

    Hence the specific heat capacity of aluminium is 908.108 J/kg/k
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