The melting point of a certain substance is , its normal boiling point is , its enthalpy of fusion i — Thermodynamics and Thermochemistry Chemistry Question
Question
The melting point of a certain substance is $70^\circ\text{C}$, its normal boiling point is $450^\circ\text{C}$, its enthalpy of fusion is $30\text{ cal/g}$, its enthalpy of vaporization is $45\text{ cal/g}$, and its specific heat is $0.215\text{ cal/g-K}$. The heat required (in cal) to convert 10 g of the substance from the solid state at $70^\circ\text{C}$ to vapour at $450^\circ\text{C}$ is
💡 Solution & Explanation
Total heat $Q = q_{\text{melt}} + q_{\text{heat liquid}} + q_{\text{boil}}$. $q_{\text{melt}} = m L_f = 10 \times 30 = 300\text{ cal}$. $q_{\text{heat}} = m c \Delta T = 10 \times 0.215 \times (450 - 70) = 2.15 \times 380 = 817\text{ cal}$. $q_{\text{boil}} = m L_v = 10 \times 45 = 450\text{ cal}$. Total $Q = 300 + 817 + 450 = 1567\text{ cal}$.