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28 April, 10:56

A 40.5 g block of an unknown metal is heated in a hot water bath to 100.0°C. When the block is placed in an insulated vessel containing 130.0 g of water at 25.0°C, the final temperature is 28.0°C. Determine the specific heat of the unknown metal. The cs for water is 4.18 J/g°C.

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  1. 28 April, 11:11
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    C = 0.5621J/g°C

    Explanation:

    Heat lost by metal = Heat gained by water

    Heat gained by water;

    H = MC ΔT

    ΔT = 28 - 25 = 3.0°C

    M = 130.0g

    C = 4.18 J/g°C

    H = 130 * 4.18 * 3 = 1639J

    Heat lost my metal;

    H = MCΔT

    ΔT = 100 - 28 = 72°C

    M = 40.5g

    C = ?

    C = H / MΔT

    C = 1639 / (40.5 * 72)

    C = 0.5621J/g°C
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