Question

What quantity of energy does it take to convert 0.700 kg ice at –20.°C to steam at 250.°C? Specific heat capacities: ice, 2.03 J/g·°C; liquid, 4.2 J/g·°C; steam, 2.0 J/g·°C; delta Hvap = 40.7 kJ/mol; delta Hfus= 6.02 kJ/mol

Answer 1

Solution

Mass of ice = m = 0.700 kg = 700 g

No. of moles of ice= n = (700 g)/ (18 g/mol) = 38.89 mole

i) Energy required to raise temperature from -20°C to 0°C

= (700 g)(2.03 J/g°C)(0-(-20°C )) = 28420 J = 28.42 kJ

ii) Energy of fusion from 0°C ice to 0°C liquid

= (38.89 mole)(6.02 kJ/mole) = 234.12 kJ

iii) Energy required to raise temperature from 0°C to 100°C

= (700 g)(4.2J/g°C)(100-0°C )) = 294000 J = 294 kJ

iv)

Energy of evaporation from 100°C liquid to 100°C vapor

= (38.89 mole)(40.7 kJ/mole) = 1582.82 kJ

v)  Energy required to raise temperature from 100°C vapor to 250°C stream

= (700 g)(2.0 J/g°C)(250-100°C )) = 210000 J = 210 kJ

Total energy required,  

= 28.42 kJ + 234.12 kJ + 294 kJ + 1582.82 kJ + 210 kJ

= 2349.36 kJ (Ans)