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2 January, 12:10

The balanced combustion reaction for C 6 H 6 is 2 C 6 H 6 (l) + 15 O 2 (g) ⟶ 12 CO 2 (g) + 6 H 2 O (l) + 6542 kJ If 8.100 g C 6 H 6 is burned and the heat produced from the burning is added to 5691 g of water at 21 ∘ C, what is the final temperature of the water?

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  1. 2 January, 14:51
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    The final temperature of water = 35.2 °C

    Explanation:

    Step 1: Data given

    Mass of C6H6 = 8.100 grams

    Mass of water = 5691 grams

    Temperature = 21 °C

    Step 2: The balanced equation

    2C6H6 (l) + 15O2 (g) ⟶ 12 CO2 (g) + 6H2O (l) + 6542 kJ

    Step 3:

    Q = m*c*ΔT.

    ⇒with Q = the heat released during this reaction (this depends on the amount of reactants used)

    ⇒ with m = the mass of the water

    ⇒with c = the "specific heat" of water = how much energy it takes to raise the temp of 1g of water by 1°C

    ⇒with ΔT = the change in the temperature of the water

    For every 2 moles of C6H6 consumed, 6542 kJ of heat is released.

    Step 4: Calculate moles for 8.100 grams

    8.100grams / 78.11 g/mol = 0.1037 mol es

    So, according to the equation, the amount of heat released is:

    (0.1037 moles / 2 moles) * (6542 kJ) = 339.2 kJ

    Step 5: Calculate the final temperature

    Q = mcΔT

    ΔT = Q / (m*c)

    T2 - T1 = Q / (m*c)

    T2 = [Q / (m*c) ] + T1 = [ (339.2 kJ) / (5691g) (0.004186 kJ/g°C) ] + 21°C = 35.2°C

    The final temperature of water = 35.2 °C
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