In an experiment, 10.0 g of ice at -20°C is converted into steam with a temperature of 110°C. How much energy is required for this process?
$\Delta$ H_vap = 2,260 J/g; $\Delta$ H_fus = 334 J/g; c_s(ice) = 2.06 J/(g . °C) ; c_s(water) = 4.18 J/(g . °C) ; C_s(steam) = 1.99 J/(g . °C) )

Question

In an experiment, 10.0 g of ice at -20°C is converted into steam with a temperature of 110°C. How much energy is required for this process?

\Delta

H_vap = 2,260 J/g;

\Delta

H_fus = 334 J/g; c_s(ice) = 2.06 J/(g . °C) ; c_s(water) = 4.18 J/(g . °C) ; C_s(steam) = 1.99 J/(g . °C) )

Quizplus · Accepted Answer

Explanation:The energy required for this process is the sum of the energy required to raise the temperature of the ice to 0°C, the energy required to melt the ice, the energy required to raise the temperature of the water to 100°C, and the energy required to vaporize the water.Energy to raise the temperature of the ice to 0°C:Q1 = m * c_s(ice) * ΔTQ1 = 10.0 g * 2.06 J/(g . °C) * (0°C - (-20°C))Q1 = 412 JEnergy to melt the ice:Q2 = m * ΔH_fusQ2 = 10.0 g * 334 J/gQ2 = 3340 JEnergy to raise the temperature of the water to 100°C:Q3 = m * c_s(water) * ΔTQ3 = 10.0 g * 4.18 J/(g . °C) * (100°C - 0°C)Q3 = 4180 JEnergy to vaporize the water:Q4 = m * ΔH_vapQ4 = 10.0 g * 2,260 J/gQ4 = 22,600 JTotal energy required:Q = Q1 + Q2 + Q3 + Q4Q = 412 J + 3340 J + 4180 J + 22,600 JQ = 30,532 JQ ≈ 30.7 kJ

In an Experiment, 10 Δ\DeltaΔ Hvap = 2,260 J/g; Δ\DeltaΔ Hfus = 334 J/g; Cs(ice) = 2

In an Experiment, 10 $\Delta$ H_vap = 2,260 J/g; $\Delta$ H_fus = 334 J/g; C_s(ice) = 2