In an experiment,10.0 g of ice at -20.0°C is converted into steam with a temperature of 110.0°C.How much energy is required for this process? [ $\Delta$ H_vap = 2260 J/g; $\Delta$ H_fus = 334 J/g; c_p (ice) = 2.06 J/(g.°C) ; c_p(water) = 4.18 J/(g.°C) ;
C_p(steam) = 1.99 J/(g.°C) ]

Question

In an experiment,10.0 g of ice at -20.0°C is converted into steam with a temperature of 110.0°C.How much energy is required for this process? [

\Delta

H_vap = 2260 J/g;

\Delta

H_fus = 334 J/g; c_p (ice) = 2.06 J/(g.°C) ; c_p(water) = 4.18 J/(g.°C) ;
C_p(steam) = 1.99 J/(g.°C) ]

Quizplus · Accepted Answer

Explanation: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, and the energy required to vaporize the water.Energy to melt the ice = 10.0 g * 334 J/g = 3340 JEnergy to raise the temperature of the water = 10.0 g * 4.18 J/(g °C) * (100°C - 0°C) = 4180 JEnergy to vaporize the water = 10.0 g * 2260 J/g = 22600 JTotal energy required = 3340 J + 4180 J + 22600 J = 30120 J = 30.1 kJTherefore, the correct answer is A) 30.7 kJ.

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

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