Question

You mix 1.19 mol of water at 80.9*C with 5.64 mol of water at 11.5*C. Determine (delta)S for the process that occurs. Assume the heat capacity of water is constant at 75.3 J/K*mol for this temperature range.

Answer to 3 significant figures.

Please show work.

(delta)S= _______J/K

Answer 1

Given moles of waterm m1 =1.19 mol, temperature T1=80.9 °C=80.9+273 K=353.9 K,

Moles of water m2=5.64 mol, temperature T2=11.5 °C=11.5+273=284.5 K.

and heat capacity C=75.3 J/K mol.

We know that,

alculation of ∆H,
∆U1 = m1xCx(T1-Tf)
∆U2 = m2xCx(Tf-T2)

Enegy balance for insulated system.
∆U = Q-W, Q = 0 W = 0
∆U = 0
∆U1+∆U2 = 0, because block is incompressible, so ∆U = ∆H
∆H1+∆H2 = 0
∆Htotal = ∆H1+∆H2 = 0

Calculation for ∆S
Find final temperature
∆H1+∆H2 = 0
m1xCx(T1-Tf) = m2xCx(Tf-T2),  
1.19(T1-Tf) = 5.64(Tf-T2 )
6.83Tf = 1.19xT1+5.64xT2
Tf = 296.59 K.

Now we can find ∆S mix,

∆S1 = m1xCxln(Tf/T1)
∆S2 = m2xCxln(Tf/T2)

∆S mix = ∆S1+∆S2 = m1xCx[ln(Tf/T1)+m2xCxln(Tf/T2)]
∆S mix = 1.19 mol*75.3 J/K mol* ln(296.59/353.9)+5.64 mol*75.3 J/K mol* ln(296.59/284.5)
∆S mix = -15.83 J/K+17.674 J/K=1.845 J/K.