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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
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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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2021-07-30T20:22:59+00:00
2021-07-30T20:22:59+00:00 1 Answers
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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