Question

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 1.25 kg of water decreased from 115 °C to 42.5 °C. Property Melting point Units °C Boiling point °C ΔHus kJ/mol Value 0 100.0 6.01 40.67 37.1 75.3 33.6 kJ/mol J/mol. "C ΔHap (s) 40( J/mol. °C J/mol. °C Co (8)

Answer 1

First find in cooling 115c to water is a which is The Heat 1.25 kg of H20 from louc. (The boiling point of - looc) - q = m.c.oT . Where 'c' for steam is 1.996 kg / kgoc me 1.25_kg :: 91 = (1.25 $9.). (1-9-9-6_kJd -(115 19 = 37.42 kJ (115 - loo €) --3-3-6_k_Jed 9 Then, find a which is heat in condensing 1.25 kg of Hao aut 100°c. The heat of vaporization of water is 2360 kJ/kq_(A 'Hvap=2260 kg ikg) 92 = m x Hrap = (1.25 kg) x (2260 kJ/kg) 79232825 kg]

the heat in H₂O from Find 93, which is cooling 1.25 kg of 100°C to 42. 5°C specific Hear of is 4.187 7 KJ /kgoc liquid water moc.at where m= 1.25 kg c=4.187 KJ /kg. "C LAT = (low-42.59°C

93 - (1.35 153).(4.187 kJ ) (100-48.5°C) - 4 КТ kao [ 93 = 300.94 kJ ! Thus, Totalo energy = 9, + 92 +93 = (37.42 kJ) + ( 28.25 KJ) + (300.94 kg] Total = 3163. 36 KI energy _ _ energy change is Total 3163. 36 KI.