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 heat is given up by $2.80 \mathrm {~kg}$ of this substance when it is cooled from 170°C to 86°C at a pressure of one atmosphere?
A)400 kJ
B)200 kJ
C)300 kJ
D)440 kJ
E)640 kJ

Question

Quizplus · Accepted Answer

The heat given up is the sum of the heat lost when the substance cools from 170°C to 150°C (in the gas phase), the heat lost when it condenses at 150°C, the heat lost when it cools from 150°C to 20°C (in the liquid phase), and the heat lost when it freezes at 20°C. The heat lost in each step is:- Cooling from 170°C to 150°C (gas phase):$$Q_1 = mc_p \Delta T = (2.80 \mathrm{~kg})(400 \mathrm{~J/kg \cdot K})(150°C - 170°C) = -33.6 \mathrm{~kJ}$$- Condensation at 150°C:$$Q_2 = mL = (2.80 \mathrm{~kg})(7.2 	imes 10^4 \mathrm{~J/kg}) = -201.6 \mathrm{~kJ}$$- Cooling from 150°C to 20°C (liquid phase):$$Q_3 = mc_p \Delta T = (2.80 \mathrm{~kg})(1000 \mathrm{~J/kg \cdot K})(20°C - 150°C) = -392 \mathrm{~kJ}$$- Freezing at 20°C:$$Q_4 = mL = (2.80 \mathrm{~kg})(3.6 	imes 10^4 \mathrm{~J/kg}) = -100.8 \mathrm{~kJ}$$The total heat given up is:$$Q = Q_1 + Q_2 + Q_3 + Q_4 = -33.6 \mathrm{~kJ} - 201.6 \mathrm{~kJ} - 392 \mathrm{~kJ} - 100.8 \mathrm{~kJ} = -728 \mathrm{~kJ}$$Therefore, the heat given up by 2.80 kg of this substance when it is cooled from 170°C to 86°C at a pressure of one atmosphere is 728 kJ.

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