Question

At 1 atm, one cup (225 grams) of water is heated from 298.15 K to the boiling temperature (373.15 K), turns completely into steam, and finally reaches a temperature of 423.15 K.

a) Compute the enthalpy change (∆H) for the entire process

b) Compute the entropy change (∆S) for the entire process

Note that i) the molecular weight of water is 18.015; ii) the heat capacity of liquid water has a roughly constant value of 75.3 J K−1 mol−1 ; iii) the heat capacity of water steam has a roughly constant value of 33.6 J K−1 mol−1 ; and iv) the enthalpy of vaporization of water is 40.65 kJ mol−1

Thank you!

Answer 1

. Ang Pressure = 1 atm wt of water = 225 grams Ti= 298.15k . Tzz 373.15K T3=423.15 K T → Te T3 Molo wt of H₂O = 18.015 Heat capacity H2O = 75-3 J/K mol Note Heat capacity 120 = 33.6J 1 kmol Enthalpy of vapourisation 40-65 kJ/mali Soly (0 E и чо Cом. ko T 298.15k le). , - Lлаш - бо Tz 373.15 eg) - Tz 423.15 (9) Enthalpy change = {msor Moles of water a 225 = 12:48 moles 184015 cs Scanned with CamScanner

DATE... PAGE....... ton (t + T2) + (Tz [3) ton M L + MGOT) + MGAT TAM = 12:48 X 4 0650 + 12:488 7513875 12.48X 33-6 X 5o St=5073127 70-487 20967 al=598 759.4 AH = 598.7 kJ - Entropy change (AS) AS = AH AT os = 598.7 423.15-298.15 AS = 598.7 - 125 Scannesh 48 CamScanner