Thus all one problem please help me I need it fast please We calculated entropy
ID: 1063776 • Letter: T
Question
Thus all one problem please help me I need it fast please We calculated entropy change delta S for ice melting. Now we have 20 g ice at initial temperature 0 X. and 200 g water at initial temperature 25 degree C. If we mix them in a calorimeter, the ice melts, and final temperature is 11 degree C. Assume that the heat capacity for calorimeter C_ calorimeter can be neglected, and molar heat capacity for water is C_p, m = 75.3 J/(mol degree C). What is the entropy change for 200 g water to cool down from initial 25 degree C to final 11degree C? What is the entropy change for 20 g ice to melt at 0 degree C? What is the entropy change for 20 g water to increase temperature from 0 degree C to the final temperature (11degree C)? What is the total entropy change? Is this process spontaneous? Why?Explanation / Answer
Entropy change for 200 gm of water, = Mass of water*Cp of water* ln (T2/T1)
Where T2= 11+273=284 ( final equilibrium temperature) and T1= 25+273=298
Entropy change for 200gm of water = 200* 4.184*ln(284/298)= -40.2662 J/K
Entropy change for 200 gm of water, = Mass of water*Cp of water* ln (T2/T1)
Where T2= 11+273=284 ( final equilibrium temperature) and T1= 25+273=298
Entropy change for 200gm of water = 200* 4.184*ln(284/298)= -40.2662 J/K
Entropy change during melting of 20 gm of water = Enthalpy of fusion/ melting temperature (K)
=335 J/g*20/273 = 24.54 J/K
Entropy change for 20 gm of liquid water for change in temperature from 0 to 11 deg.c , = Mass of water*Cp of water* ln (T2/T1)
Where T2= 11+273=284 ( final equilibrium temperature) and T1= 0+273=273K
Entropy change for 20gm of water = 20* 4.184*ln(284/273)= 3.305 J/K
Total entropy change for 20 gm of water = 24.54 +3.305 =27.845 J/K
Entropy change of system = -40.2662+27.845=-12.4212 J/k
Entropy change is an indication of spontaneity of this process. Entropy change if is +ve the process is spontaneous. In the present case, it is –ve and the process is non-spontaneous.
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