A substance has a melting point of 20°C and a heat of fusion of $3.6 \times 10^4 \mathrm {~J} / \mathrm { kg }$ The boiling point is $150 ^ { \circ } \mathrm { C }$ and the heat of vaporization is $7.2 \times 10^4 \mathrm {~J} / \mathrm { kg }$ at a pressure of one atmosphere.The specific heats for the solid,liquid,and gaseous phases are 600 J/kg ∙ K (solid),1000 J/kg ∙ K (liquid),and 400 J/kg ∙ K (gaseous).How much thermal energy is emitted by $2.80 \mathrm {~kg}$ of this substance when it is cooled from 170°C to 86°C at a pressure of one atmosphere?

Question

Quizplus · Accepted Answer

First, we need to calculate the energy required to cool the substance from 170°C to 20°C (its melting point), and then to cool it further to 86°C.

Energy required to cool from 170°C to 20°C:
- Phase change from gas to liquid: q = m × ΔHvap = 2.8 kg × 7.2 × 10^4 J/kg = 2.016 × 10^5 J 
- Cooling liquid from boiling point to melting point: q = m × Cp × ΔT = 2.8 kg × 1000 J/kg ∙ K × (150°C - 20°C) = 308 × 10^3 J 
- Phase change from liquid to solid: q = m × ΔHfusion = 2.8 kg × 3.6 × 10^4 J/kg = 1.008 × 10^5 J 
Total energy required to cool from 170°C to 20°C: 2.016 × 10^5 J + 308 × 10^3 J + 1.008 × 10^5 J = 4.032 × 10^5 J

Energy required to cool from 20°C to 86°C:
- Cooling solid from melting point to 86°C: q = m × Cp × ΔT = 2.8 kg × 600 J/kg ∙ K × (86°C - 20°C) = 1.3728 × 10^5 J

Total energy emitted by the substance: 4.032 × 10^5 J + 1.3728 × 10^5 J = 5.4048 × 10^5 J ≈ 400 kJ.

Therefore, the best choice is A) 400 kJ, which is the closest to the calculated value.

A Substance Has a Melting Point of 20°C and a Heat