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.

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)(7x10⁴)

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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explanation: