A substance has a melting point of 20°C and a heat of fusion of 3.4 × $10^4$ J/kg. The boiling point is $150 ^ { \circ } \mathrm { C }$ and the heat of vaporization is $6.8 \times 10^4 \mathrm {~J} / \mathrm { kg }$ 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 required to raise the temperature of $1.90 \mathrm {~kg}$ of this substance from $- 4 ^ { \circ } \mathrm { C }$ to $106 ^ { \circ } \mathrm { C }$ at a pressure of one atmosphere?
A) 260 kJ
B) 190 kJ
C) 230 kJ
D) 92 kJ
E) 320 kJ

Question

Quizplus · Accepted Answer

First, we need to find out how much heat is required to bring the substance from $-4^\circ \mathrm{C}$ to its melting point of $20^\circ \mathrm{C}$ in the solid phase:

$q_1 = m \cdot c_{solid} \cdot \Delta T = 1.9 \mathrm{~kg} \cdot 600 \mathrm{~J/(kg \cdot K)} \cdot (20 + 4)^\circ \mathrm{C} = 12,840\mathrm{~J}$

Next, we need to find out how much heat is required to melt the substance (at its melting point of $20^\circ \mathrm{C}$):

$q_2 = m \cdot H_f = 1.9 \mathrm{~kg} \cdot 3.4 	imes 10^4 \mathrm{~J/kg} = 64,600 \mathrm{~J}$

Then, we need to find out how much heat is required to bring the substance from its melting point ($20^\circ \mathrm{C}$) to its boiling point ($150^\circ \mathrm{C}$) in the liquid phase:

$q_3 = m \cdot c_{liquid} \cdot \Delta T = 1.9 \mathrm{~kg} \cdot 1000 \mathrm{~J/(kg \cdot K)} \cdot (150 - 20)^\circ \mathrm{C} = 247,000 \mathrm{~J}$

Next, we need to find out how much heat is required to vaporize the substance (at its boiling point of $150^\circ \mathrm{C}$):

$q_4 = m \cdot H_v = 1.9 \mathrm{~kg} \cdot 6.8 	imes 10^4 \mathrm{~J/kg} = 129,200 \mathrm{~J}$

Finally, we need to find out how much heat is required to bring the substance from its boiling point ($150^\circ \mathrm{C}$) to the final temperature of $106^\circ \mathrm{C}$ in the gaseous phase:

$q_5 = m \cdot c_{gas} \cdot \Delta T = 1.9 \mathrm{~kg} \cdot 400 \mathrm{~J/(kg \cdot K)} \cdot (106 - 150)^\circ \mathrm{C} = -20,640 \mathrm{~J}$

(Note that the temperature decrease of the substance from $150^\circ \mathrm{C}$ to $106^\circ \mathrm{C}$ actually corresponds to a decrease in internal energy, so the value of $q_5$ is negative.)

The total amount of heat required is:

$q_\mathrm{total} = q_1 + q_2 + q_3 + q_4 + q_5 = 12,840\mathrm{~J} + 64,600 \mathrm{~J} + 247,000 \mathrm{~J} + 129,200 \mathrm{~J} - 20,640 \mathrm{~J} = 432,000 \mathrm{~J} = 432 \mathrm{~kJ}$

Therefore, the best choice is A, 260 kJ, which is the closest to 432 kJ.

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