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if 8.100 grams of C6H6 is burned and heat is produced from the burning is added to 5691 grams of water at 21 degrees celsius what is the final temp of the water?
2C6H6+15O2=12CO2+6H2O+6524kj

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  1. 2C6H6 + 15O2 ==> 12CO2 + 6H2O

    Then heat from 8.1 g will be
    6524 kJ x (8.100g/2*molar mass C6H6) = y kJ.
    Substitute y into
    y = mass H2O x specific heat H2O x (Tfinal-Tinitial). Tf is the only unknown.

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  2. C6H12 = 6x12 + 6x1 = 78.
    The equation indicates that 2x78 = 156g benzene will produce 6542kJ.
    Using proportions you can then calculate that
    x/6542kJ = 8.1g / 156g
    x = 339.7kJ = 339700J.

    heat = mass x ΔT x 4.18J/g°
    ΔT = 339700J / (5691g x 4.18J/g°) = 14.3°

    final temp = 21 + 14.3° = 35.3°C

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