How much energy is required to heat 36.0 g H₂O from a liquid at 65 °C to a gas at 115 °C? The following physical data may be useful: Δ_vapH = 40.7 kJ mol^-1
C_liq = 4.18 J g^-1 °C^-1
C_gas= 2.01 J g^-1 °C^-1
C_sol = 2.09 J g^-1 °C^-1
T_melting= 0 ^∘C
T_boiling = 100 ^∘C

Question

Quizplus · Accepted Answer

To calculate the total energy required, we need to consider three steps: heating the liquid water from 65°C to 100°C, vaporizing the water at 100°C, and then heating the steam from 100°C to 115°C. 1. Heating liquid water from 65°C to 100°C:q1 = m * C_liq * ΔT = 36.0 g * 4.18 J/g°C * (100°C - 65°C) = 5274 J = 5.274 kJ2. Vaporizing the water at 100°C:First, convert grams to moles: 36.0 g / 18.015 g/mol = 2.0 molq2 = n * Δ_vapH = 2.0 mol * 40.7 kJ/mol = 81.4 kJ3. Heating steam from 100°C to 115°C:q3 = m * C_gas * ΔT = 36.0 g * 2.01 J/g°C * (115°C - 100°C) = 1085.4 J = 1.0854 kJTotal energy = q1 + q2 + q3 = 5.274 kJ + 81.4 kJ + 1.0854 kJ = 87.7594 kJ ≈ 87.7 kJ

How Much Energy Is Required to Heat 36