A substan e has a melting point of 20◦C and a heat of fusion of 3.5 × 104 J/kg. The boiling point is 150◦C and the heat of vaporization is 7

Question

A substan e has a melting point of 20◦C and a heat of fusion of 3.5 × 104 J/kg. The boiling point is 150◦C and the heat of vaporization is 7.0 × 104 J/kg at a pressure of 1.0 atm. The spe i heats for the solid, liquid, and gaseous phases are 600 J/(kg.K), 1000 J/(kg.K), and 400 J/(kg.K), respe tively. The quantity of heat given up by 0.50 kg of the substan e when it is ooled from 170◦C to 88◦C, at a pressure of 1.0 atmosphere, is losest to (1) 70 kJ (2) 14 kJ (3) 21 kJ (4) 30 kJ (5) None of the above.

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Eirian 4 years 2021-07-30T20:22:59+00:00 1 Answers 105 views 0

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    2021-07-30T20:24:32+00:00

    Answer: Option (1) 70 kJ

    Explanation:

    Given that;

    melting point of the substance T₁ is 20°C
    , Boiling point of the substance T₂ is 150°C
    , heat of fusion L₁ is 3.5 x 10⁴ J/kg
    , heat of vaporization L₂ is 7 x 10⁴ J/kg
    , Specific heat in solid state C₁ is 600 J/kg.K
    , Specific heat in liquid state C₂ is 1000 J/kg.K, Specific heat in gaseous state C₃ is 400 J/(kg.K)
    , Mass of the substance m is 0.5 kg
    , Initial temperature of the substance T₃ is 170°C
    , Final temperature of the substance  T₄ is 88°C
    .

    Now  Heat given up by the substance to reach boiling point 150°C is

    Q₁ = mC₃(T₃ – T₂)

    Q₁ = (0.5)(400)(170 – 150)

    Q₁ = 4000 J

    Heat given up by the substance to turn into liquid from gaseous state at 150°C

    Q₂ = mL₂

    Q₂ = (0.5)(7×10⁴)

    Q₂ = 35000 J

    Heat given up by the substance to reach 88°C in liquid state from 150°C

    Q₃ = mC₂(T₂ – T₄)

    Q₃ = (0.5)(1000)(150 – 88)

    Q₃ = 31000 J

    Total heat given up by the substance

    Q = Q₁ + Q₂ + Q₃

    Q = 4000 + 35000 + 31000

    Q = 70000 J


    Q = 70 kJ

    Therefore The quantity of heat given up by 0.50 kg of the substance when it is cooled from 170◦C to 88◦C, at a pressure of 1.0 atmosphere, is closest to 70 kJ

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