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8 December, 11:40

If 55.0 mL of ethanol (density = 0.789 g/mL) initially at 8.0 ∘C is mixed with 55.0 mL of water (density = 1.0 g/mL) initially at 28.4 ∘C in an insulated beaker, what is the final temperature of the mixture, assuming that no heat is lost? (CEtOH=2.42J / (g⋅∘C).) Express the temperature in degrees Celsius to three significant figures.

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  1. 8 December, 12:01
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    T = 45.509 °C

    Explanation:

    Q = mCΔT

    EtOH:

    ∴ m EtOH = (55.0 mL) * (0.789 g/mL) = 43.395 g

    ∴ CEtOH = 2.42J/g°C

    ∴ T1 = 8.0 °C

    H2O:

    ∴ m H2O = (55.0mL) * (1.0 g/mL) = 55 g

    ∴ C H2O = 4.186 J/g°C

    ∴ T1 = 28.4 °C

    final temperature of the mixture (T):

    ⇒ Q EtOH = Q H2O

    ⇒ (43.395 g) (2.42 J/g°C) (T - 8.0°C) = (55.0 g) (4.186 J/g°C) (T - 28.4°C)

    ⇒ 105.016 T - 840.127 = 230.23 T - 6538.532

    ⇒ 6538.532 - 840.127 = 230.23 T - 105.016 T

    ⇒ 5698.405J = (125.214 J/°C) (T)

    ⇒ T = 45.509 °C
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