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29 April, 12:34

A 77.5 g sample of an unknown solid is heated to 62.5°C and placed into a calorimeter containing 93 g of water at 23.3°C. If the final temperature of the solid sample and the water is 26.2°C, what is the specific heat of the solid? Assuming no heat loss.

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  1. 29 April, 15:36
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    Specific heat of solid is 0.401J/g°C

    Explanation:

    In this experiment, the heat losed for the unknown solid is equal to heat gained for the water.

    Using calorimeter equation:

    Q = m*C*ΔT

    Where Q is heat, m is mass of substance, C is specific heat and ΔT is change in temperature.

    It is possible to write:

    m[Solid] * C[Solid] * ΔT[Solid] = m[Water] * C[Water] * ΔT[Water]

    Replacing:

    77.5g * C[Solid] * (62.5°C - 26.2°C) = 93g * 4.184J/g°C * (26.2°C - 23.3°C)

    77.5g * C[Solid] * (62.5°C - 26.2°C) = 93g * 4.184J/g°C * (26.2°C - 23.3°C)

    2813.25g°C * C[Solid] = 1128.4J

    C[Solid] = 0.401 J/g°C

    Specific heat of solid is 0.401J/g°C
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