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22 April, 00:31

A 140.0-g sample of water at 25.0°c is mixed with 111.7 g of a certain metal at 100.0°c. after thermal equilibrium is established, the (final) temperature of the mixture is 29.6°c. what is the specific heat capacity of the metal, assuming it is constant over the temperature range concerned?

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  1. 22 April, 02:42
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    Mass of water = 140.0 g

    Initial temperature of water = 25.0°C

    Mass of a certain metal = 111.7 g

    Initial temperature of metal = 100.0°C

    Final temperature of water and metal = 29.6°C

    Since the metal is at a higher initial temperature it will lose heat and the water having a lower initial temperature will gain heat.

    Thus, heat lost by metal = Heat gained by water

    formula: (mass metal) (initialT - finalT) (Cp metal) = (mass water) (finalT - initalT) (Cp water)

    After plugging in the given data we get,

    (111.7g) (100 °C - 29.6°C) (Cp metal) = (140.0g) (29.6°C-25.0°C) (4.184 J/g°C)

    (111.7g) (70.4°C) (Cp metal) = (140.0g) (4.6°C) (4.184 J/g°C)

    (7863.7 g°C) (Cp metal) = 2694.5 J

    (Cp metal) = 2694.5 J / 7863.7 g°C

    Thus, Cp of metal = 0.3427 J/g°C
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