Calculate the change in entropy when 50 g ofwater at 80 °C is poured into 100 g

Question

Calculate the change in entropy when 50 g ofwater at 80 °C is poured into 100 g of water at 10 °C in aninsulated vessel. Given that Cm,p = 75.5J´K-1´mol-1.

here the pressure seems to be constant, so variation of entropywith temperature for constant pressure process is given by:

dS = Cp d(lnT)

now integrating the entropy from S1 to S2 and temperature fromT1 to T2

we get S2 = S1 + integartion from limits T1 to T2 of Cpd(lnT)

so first we need to find the final temperature of the water.

here when 50 g of water at 80 °C is poured into 100 g ofwater at 10 °C in an insulated vessel let final temp be t

so heat lost by 50 gm water = mcT

                                          = 50*1*(80+273.15-t)

and heat gained by 100 gm water = 100*1*(t-10-273.15)

so now both are equal so

50(80+273.15-t) = 100(t-10-273.15)

4000 -50t+ 50*273.15 = -100*273.15+100t - 1000

3000 + 150*273.15 = 50t

so t = 879.45 K

so final temp is 879.45 K

so now entropy change dS = dS1 + dS2

now dS1 = -75.5 ln(879.45 -80-273.15)

             = -473.07

now dS2 = 75.5ln(879.45-10-273.15)

             = 482.50

total dS = 9.4311 J/K/mol

Explanation / Answer

here for we calculate heat energy for water both thecold, hot conditionn by differing their intial and finaltemperatures. later we can calculate entropy change for both hotand cold water systems. Later add both of them to get final changein entropy for system.    equations used are: Q = m.s.T                               s = CplndT

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