Question

5) How much heat is required to convert 0.05kg of water at 298 K to super-steam at 423 K. The boiling point of water is 373 K. Cm [H2O (l)] = 75.4 J/(mol. ℃) Cm [H2O (g)] = 33.6 J/(mol. ℃) ΔHvap = 40.67 75.4 J/mol

Answer 1

molar heat capacity (cm) = 2 O= heat required to change the substance on giving the temperature m= moles of the substance DT = Diff in temperature or change in temperature - When we heat the substance Now we have to calculate the heat required to convert water (298k) to super steam at (423 k) This firstly the water (298 k) heated to its - Boiling Point (373k) They heat required Q= cm (Hgo, l) x mx DT Cm CHgO, 1) = 75.45| moloc m= 0.05 kg x1000 = 25 moles к 18 м сlаr mам у LOT = 373 – 298 – 75% ! water =189) Diff in oc & in k is name ie 100°c - 25% = 75° So 9,= 7514 x 25 x 75-15708133 J SD) & OM vap = 4067754 : Jl mor CS Scanned with = 40677.54 J/mol CamScanner

state chang is latent heat of vaporisation ive - Chang from water to steam ü = 40677.54 Il mol from during the heat required o Now the stam (373k) 25 so the latent head of vaporisation for 25 meles of water is Q₂ = 40677.54 x 25 J Qle = 112993.167 J - Now the steam (373k) have to convert to super steam (423K) then the heat required Q- Cm (420,g) * mx ot = 336 x 25 x (423 - 373) = 33.6 x 25 x 50 9 - 466617 J Thus total heat required = head leeg conuert H90 (298 k) to Boil-it ( 373 k) + heat required to convert Boiled water to 1 steam (Phase change from liquid to gas) + heat required to convert steam (373) to 423 k a = 0,+ O₂ +Q3 E15708.33 + 1)299 3.167 + 4666.7 51333681197 J Scanned with CamScanner = 133.368 kJ