2017 Fal Final Test Date: Name ashecaryc. 1) In exp. 5, we calculated entropy ch
ID: 562167 • Letter: 2
Question
2017 Fal Final Test Date: Name ashecaryc. 1) In exp. 5, we calculated entropy change AS for ice melting Now we have 18 g ice at initial temperature 0 , and 180 g water at initial temperature 25 "C. If we mix them in a calorimeter, the ice melts, and final temperature is 1 1 Assume that the heat capacity for calorimeter Calorimeter can be neglected, and molar heat capacity for water is C-75.3 J / (mol °C). i) What is the entropy change for 180 g water to cool down from initial 25 °C to final 11 °C? ii) What is the entropy change for 18 g ice to melt at 0 °C? ii) What is the entropy change for 18 g water to increase temperature from 0 °C to the final temperature (11 °C)? iv) What is the total entropy change? v) Is this process spontaneous? Why?Explanation / Answer
i)
dS = mass*Cp*ln(T2/T1)
Cp = 4.184 J/gK
T2 = 11°C = 11+273 = 284
T1 = 25°C = 25+273 = 298
dS = 180*4.184 *ln(284/4.184 )
dS = 3176.4 J/K
ii)
entropy change --> 18 g melting at 0°C
dS = n*dSfussion
dS = mass*dHfussion/T = 18*334/273
dS = 22.02 J/K
iii)
dS = mass*Cp*ln(T2/T1)
Cp = 4.184 J/gK
T1 = 11°C = 11+273 = 284
T2 = 0°C = 0+273 = 273
dS = 18*4.184 *ln(284/273)
dS = 2.9750 J/K
iv)
Total entropy --> addition of all
dS = 2.9750 +22.02 +3176.4
dS = 3201.395
v)
entropy increases, therefore it must be spontaneous
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