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1 November, 06:24

If 5.400 g of c6h6 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. 1 November, 06:39
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    Actually, we will use this formula:

    Q = m. C.ΔT

    when Q is the heat released during the reaction.

    and m is the mass of water.

    and C is the specific heat of water

    and ΔT is the change of the temperature of the water.

    So according to the balanced equation for this reaction:

    2C6H6 (l) + 15O2 (g) → 12CO2 (g) + 6H2O (l) + 6542 KJ

    So when 2 moles of C6H6 released 6542 KJ of heat, So how many moles of the 5.4 g

    no of moles of C6H6 = 5.4 / molar mass of C6H6

    = 5.4g/78.11 g/mol = 0.069 moles

    so 2 moles of C6H6 released → 6542

    ∴ 0.069 moles of C6H6 will release → X

    ∴Q = (0.069 * 6542) / 2 = 226 KJ

    So by substitution in Q = m. c.ΔT to get ΔT

    ∴Δ T = Q / (m. c) = 226 / (5691*0.004186) = 9.496

    when we have Ti

    ΔT=Tf-Ti

    ∴Tf = 9.496 + 21 = 30.5 °C
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