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3 November, 10:10

A 75 g sample of a substance is placed in boiling water until its temperature reaches 100 C. It is then immediately transferred to a calorimeter containing 100 g of water at 24.4 C. The final temperature of the sample and the water is 34.9 C. What is the specific heat of the sample?

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  1. 3 November, 12:40
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    specific heat of sample = 0.898 j/g.°C

    Explanation:

    Given dа ta:

    Mass of sample = 75 g

    Initial temperature of water = 100°C

    Mass of water = 100 g

    Initial temperature of water = 24.4°C

    Final temperature of sample ans water = 34.9°C

    Specific heat of sample = ?

    Solution:

    ΔT (water) = 34.9°C - 24.4°C

    ΔT (water) = 10.5°C

    ΔT (sample) = 34.9°C - 100°C

    ΔT (sample) = - 65.1°C

    -Q (sample) = Q (water)

    -m*c*ΔT = m*c*ΔT

    -[75 g * c * - 65.1°C] = 100 g * 4.18 j/g.°C * 10.5°C

    -[-4882.5 g.°C * c = 4389 j

    4882.5 g.°C * c = 4389 j

    c = 4389 j / 4882.5 g.°C

    c = 0.898 j/g.°C
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