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10 April, 01:43

A 1.00*1 0 2 g aluminum block at 100.0°C is placed in 1.00*1 0 2 g of water at 10.0°C. The final temperature of the mixture is 26.0°C. What is the specific heat of the aluminum?

CA = (-mBCB∆TB) / mA∆TA

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  1. 10 April, 02:25
    0
    The answer to your question is Ca = 0.904 J/g°C

    Explanation:

    Data

    Aluminum Water

    mass 1 x10² g 1 x 10² g

    Temperature 1 100°C 26°C

    Temperature 2 10°C 10°C

    Specific heat? 4.182 J/g°C

    Formula

    Aluminum Water

    -mCa (T2 - T1) = mCw (T2 - T1)

    -Solve for Ca

    Ca = [mCw (T2 - T1) ] / - m (T2 - T1)

    -Substitution

    Ca = [1 x 10² x 4.182 x (26 - 10] / - 1 x10² (26 - 100)

    -Simplification

    Ca = [418.2 (16) ] / - 1 x 10² (-74)

    Ca = 6691.2 / 7400

    -Result

    Ca = 0.904 J/g°C
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