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 = 2260 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 = 2260 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

The correct answer is A.The energy required for this process is the sum of the energy required to melt the ice, the energy required to raise the temperature of the water from 0°C to 100°C, the energy required to vaporize the water, and the energy required to raise the temperature of the steam from 100°C to 110°C.The energy required to melt the ice is:Q1 = m * ΔHfus = 10.0 g * 334 J/g = 3340 JThe energy required to raise the temperature of the water from 0°C to 100°C is:Q2 = m * c * ΔT = 10.0 g * 4.18 J/(g · °C) * 100°C = 4180 JThe energy required to vaporize the water is:Q3 = m * ΔHvap = 10.0 g * 2260 J/g = 22600 JThe energy required to raise the temperature of the steam from 100°C to 110°C is:Q4 = m * c * ΔT = 10.0 g * 1.99 J/(g · °C) * 10°C = 199 JThe total energy required for this process is:Q = Q1 + Q2 + Q3 + Q4 = 3340 J + 4180 J + 22600 J + 199 J = 30319 J ≈ 30.3 kJ

In an Experiment, 10 Δ\DeltaΔ Hvap = 2260 J/g; Δ\DeltaΔ Hfus = 334 J/g; Cs(ice) = 2

In an Experiment, 10 $\Delta$ H_vap = 2260 J/g; $\Delta$ H_fus = 334 J/g; C_s(ice) = 2