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4 July, 06:52

If 2500. J of energy are added to 120. g of benzene at 30. degrees C, what will be its final temperature?

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  1. 4 July, 08:05
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    When heat (q) is absorbed by "m" grams of a substance then the change in temperature is given as,

    q = m Cp (T₂ - T₁) - - - (1)

    where;

    Cp = Specific Heat

    Specific heat of Benzene is 1.72 J/g.°C

    Now,

    Solving equation 1,

    (T₂ - T₁) = q / m Cp

    Putting values,

    (T₂ - 30 °C) = 2500 J : (120 g * 1.72 J/g.°C)

    (T₂ - 30 °C) = 2500 J : (206 J/°C)

    (T₂ - 30 °C) = 12.13 °C

    T₂ = 12.13 °C + 30 °C

    T₂ = 42.13 °C
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