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 thermal energy is required to raise the temperature of $1.90 \mathrm {~kg}$ this substance from $- 4 ^ { \circ } \mathrm { C }$ to $106 ^ { \circ } \mathrm { C }$ at a pressure of one atmosphere?

Question

Quizplus · Accepted Answer

To calculate the total thermal energy required, we need to consider the energy needed for each phase change and temperature change:  1. Heating the solid from -4°C to 20°C: $Q_1 = m \cdot c_{\text{solid}} \cdot \Delta T = 1.90 \, \text{kg} \cdot 600 \, \text{J/kg} \cdot (20 - (-4)) \, \text{K} = 1.90 \cdot 600 \cdot 24 = 27360 \, \text{J}$.  2. Melting the solid at 20°C: $Q_2 = m \cdot \text{heat of fusion} = 1.90 \, \text{kg} \cdot 3.4 \times 10^4 \, \text{J/kg} = 64600 \, \text{J}$.  3. Heating the liquid from 20°C to 106°C: $Q_3 = m \cdot c_{\text{liquid}} \cdot \Delta T = 1.90 \, \text{kg} \cdot 1000 \, \text{J/kg} \cdot (106 - 20) \, \text{K} = 1.90 \cdot 1000 \cdot 86 = 163400 \, \text{J}$.  Total energy $Q = Q_1 + Q_2 + Q_3 = 27360 + 64600 + 163400 = 255360 \, \text{J} = 255.36 \, \text{kJ}$.  Rounding to the nearest choice, the closest is 260 kJ.

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