A pot of water with 258 g of water is boiling. At 100 o C how much energy (in kJ

Question

A pot of water with 258 g of water is boiling. At 100oC how much energy (in kJ) will need to be added until all of the water converts to steam? Hvap = 40.7 kJ/mol

How much energy must be removed to convert 9.73 g of water at 20.9 oC to an ice cube at -12.0 oC? Give your answer as a positive value in Joules since only positive energy can be removed..

cwater = 4.184 J/goC, cice = 2.03 J/goC, Hfus, water = 6.01 kJ/mol

A substance has a vapor pressure of 0.244 atm at 347 K. What is the normal boiling point of the substance in kelvin? Normal boiling is when vapor pressure = 1 atm or 760 mmHg. Hvap = 12.3 kJ/mol

Explanation / Answer

1) Molar mass of water = 18 g/mole

Thus, moles of water in 258 g of it = mass/molar mass of water = 258/18 = 14.334

Energy required to be added to convert to steam = moles of water*enthalpy of vapourisation = 14.334*40.7 = 583.367 kJ

2) moles of water in 9.73 g of it = mass/molar mass of water = 9.73/18 - 0.541

energy released when 9,.73 g of watre at 20.9 0C is brought to 0 0C = mass of water*specific heat capacity of water*change in temperature = 9.73*4.184*(20.9-0) = 850.845 J = 0.851 kJ..............(A)

energy released when water at 0 0C is converted to ice at 0 0C = moles of water*enthalpy of fusion = 0.541*6.01 = 3.249 kJ .................(B)

energy released when ice at 0 0C is freezed to -12 0C = mass of ice*specific heat capacity of ice = 9.73*2.03 = 19.752 J = 0.0198 kJ..............(C)

Thus, total heat required to be removed = (A) + (B) + (C) = 4.1198 kJ

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