Question

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.350 kg of water decreased from 117 °C to 23.0 C.

Units Property Melting point Boiling point Value 0 °C OC 100.0 6.01 40.67 Alfus Alvap kJ/mol kJ/mol 37.1 co(s) Cp (1) J/mol ·°C J/mol ·°C 75.3 Cp (8) 33.6 J/mol · °C

Answer 1

Energy change will be in negative as temperature is decreasing from 117 to 23 ⁰C. In this total process there are three types of energy changes. 1. At vapor stage 117-100⁰C         2. Energy change at condensation at 100⁰C    3. At liquid stage 100-23⁰C. Please find the attached image file for calculative steps.

So the steps to follow in energy changes calculations are - ) 117°C- 100°c in vapour phase 2 Enthalpy Changes in vapour - warter at 10000 3 100°C -23°c in ligued phase Iso m = 0. 350 kg a 350 gm n or mol 39350/18.02'3 19.42 mol. for step 1 AT' =( 100 - 11701 -17°C we know aamcall 3350 x 33.6 x(17) ē - 199,920 ] for Step 2 AH = NAH & condensation = 19.42 x F 40670 ) .-789811.4) for Step-3 so 2 (born 23 - 100) - - 77 °C 2 m CAT . 350 x 75.3 x (770) - 20293357 Total energy changes = 1+ 2+ 3 -(199920 + 789811.4 +2029335) 5 = - 3019066 Jo a 3019.066ka