Question

4) (12 marks) A bowl of warm water (T = 45.0°C) is placed in a large freezer which is maintained at -25.0°C. The mass of the water is 0.500 kg. Liquid water has Cp = 4184 J/(K•kg). The latent heat of freezing is ly = -3.35 x 109 J/kg (the negative sign is because it's the opposite of the latent heat of fusion). For ice, Cp = 2301 J/(K.kg). (a) Calculate AS water, ASgurr, and AS univ for the water cooling. (b) Calculate A Swater, ASsurr, and A Sunid for the water freezing into ice. (c) Calculate AS water, ASsurr, and AS univ for the ice cooling to the freezer's ambient temperature.
Please explain the coversions such as converting the Cp from J/k.kg. Because i looked at the previously posted questions it seems wrong. I think coverting it to J/k.mol should be multiply by 18/1000 (for water,ice) but they are doing 1000:18, dont know why. Thank you.

Answer 1

Know thut we a n entroe1 Change infiniesimat wuten Coui a) tamCp d T of tu Specis c Hat wher water hence, T mCp In water 4194 /CK.K) T 3 (233 45 T; 3 1 K so 218 J/K 319.196 4I 2 AS wate Capp AS Wwen - 319. 2

SwU 73 25 WWOu mep CT:- T -O. 5 41?4 » 45 J 0.S 4194 4 5 ASSW 223-25) 39. 59674 2 "/n 379. 6 J/ Capppon Sswn Swater 2-31.11 64 972) 339.56 /MCafpron) 60 4

neuzi ompenatwre T 23 enchan , ASken 3.35 x 10' 0.S A Saen 23 G13. S S 311 3 6 ASsuter 7 t Sunniig heat added to asurn (0.53.35 Hmperatn mbiwt (273-25) k 24 K o. S 3.35» 24 675. 40 322 5 J/n

G5 4 ASswn hena 0.SM 3,35 A Suniv 248 23 4Sw /K G18 5o 1 2 2 3/(ftoo GI Crli fon ce Tf In mCe (273- 25) 2 301 3. . K iren, ce of ice 241 ASica 0.S2301» In a33 - 0. 435 239 J/K - \0. 5 J)K (apprun

Summnd'n3 fu ter e erutry fon Now T 293-25)n to 24 K th at enchnn 4aSwPp m op (Ti- Te mCp CT-T) Ts 0.S 230 » L5 248 1s9978 226 7/K 15. 99 T/KCappr ASsurr Ssunr ASuni 9338226- | 0. 417529 1) T Suniv 5.4? 0 2. G81 J/ J/ nPPron S.48 aSuriv