Question

QUESTION 5 An ice cube at 0.00 C with a mass of 4.52 g is placed into 56 g of water, initially at 23°C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the lo in molted? (Specific heat of water in 4.184J/g °C)

Answer 1

Aas) we know that: heat required to mettice - heat released from water (warm) 9,792 = -92 -@ where a = heat that is absorbed by Ice at ooc 92. heat that is abscobed by wate at ooc 93= heat released by warm water Эives: рақ ea & = 4-529 Molos mass gle = 18.01528 g/mol Moles glce = Mass gice Molae mass gice = 4.529 18.01528 Almal - 0.250898126 mol H2O a = n Allfusion - 0 whese na nog moles gie (water) Alfusion = molas heat of fusion of water =6.01 kalmal

9,- 0.250898126 molx 6.01 kt/mor = 1.50789774 kJ = 1.58789774 x18 J 92= MCAT (2) m= mass of lie (water) = 44.52g C= Specific heat g water = 4.184 Ilgºc Ate Fihal temperature - Initial temperature = Tf - oo (Is-03°C 93= mcat - (3) where m= mass of water = 55g = Specific heat of water = 4.1847/9°C AT- TF - 23°C. ={ts - 23) °c substituting the values in ean. @ 1:50789124 xo® + 4.629x 4.1841 x(-0)% -- (554x41845 x TS-23) 8

1.50789774 x103 + 18.91168 Cuf-0) = - 230:12 (76-23) 1-50789774x10? + 18. 91168 Tf -0 = - 230-1274 – 5292.76 18.911685f + 230.125 = -5292:76 -1.50789774x10° 249.0316854 = -6800.65774 Te -6800.65774 249.03168 = -27.3084041°C = -27.3°C