Question

A piece of metal at 80

Answer 1

Tm = 80 deg. C
Vw = 1.2 litres
Tw = 72 deg. C
Tf = 76 deg. C
?w = 1000 kg/m^3 = 1 kg/lt

Solve for metal mass and its specific heat,
Mm*cm*(Tm - Tf) = Mw*cw*(Tf-Tw)
Cm(Tm - Tf) = ?w*Vw*cw*(Tf-Tw)
Cm = ?w*Vw*cw*(Tf-Tw) / (Tm - Tf)
Cm = 1*1.2*4.18*(76-72) / (80-76)
Cm = 5.016 kJ/K

Entropy change of water
?Sw = mw*cw*ln(Tf/Tw)
?Sw = 1.2*4.18*ln(349/345)
?Sw = 0.0578219 kJ/K


Entropy change of metal
?Sm = Cm*ln(Tf/Tm)
?Sm = 5.016 ln(349/353)
?Sm = -0.057163 kJ/K

Overall entropy change,
?S = ?Sw + ?Sm
?S = 0.0578219 - 0.057163
?S = 6.589659*10^-4 kJ/k
?S = 0.65897 J/K