A substance has a melting point of 20C and a heat of fusion of 3.5×10 4 J/kg. Th

Question

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

Explanation / Answer

melting point of the substance = 20 degree C

heat of fusion = 3.5 * 104 J/kg

boiling point = 150 degree C

heat of vaporisation = 7 * 10^4 J/kg

specific heat of solid state = 600 J / (kg.K)

specific heat of liquid state = 1000 J / (kg.K)

specific heat of gaseous state = 400 J / (kg.K)

at 170 degree substance will be at gaseous state so

heat released on cooling to 150 degree where it'll come to liquid state

heat = mass * specific heat * change in temperature

heat = 0.5 * 400 * (170 - 150)

latent heat = mass * heat of vaporisation

latent heat = 0.5 * 7 * 10^4

heat released on cooling from 150 degree C to 88 degree C

heat = 0.5 * 1000 * (150 - 88)

total heat =  0.5 * 400 * (170 - 150) + 0.5 * 7 * 10^4 + 0.5 * 1000 * (150 - 88)

total heat = 70000 J ot 70 kJ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.