Question

A piece of metal at 80

Answer #1

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

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