A wafer of silicon of mass, 0.2 kg, is initially at 800 o C. It is kept in a rig

Question

A wafer of silicon of mass, 0.2 kg, is initially at 800 oC. It is kept in a rigid insulated chamber of volume 0.3 m3 containing air that is initially at 1 atmosphere and 25 oC. The isolated system composed of wafer and air reaches final temperature of Tf. Neglecting the mass of the chamber material and assuming constant specific heats for air and silicon, determine the total entropy generation. Use the following properties: Silicon, c=0.712 kJ/kgK; Air: R=0.287 kJ/kgK, cp=1 kJ/kgK. (25 points)

Use the 1st law to find temperature.

Use the second law and the expression for entropy change in air (?Sair) and silicon (?SSi).

Obtain the total entropy change of the combined system of air and silicon; ?Stotal = (?Sair) + ?SSi.

Explanation / Answer

1 atm = 101.325 kPa

Mass of air m = PV / RT

= 101.325*0.3 / (0.287*(25 + 273))

= 0.3554 kg

Heat lost by wafer = Heat gained by air

[m*Cp*(delta T)]_wafer = [m*Cp*(delta T)]_air

0.2*0.712*(800 - Tf) = 0.3554*1*(Tf - 25)

Tf = 246.8 deg C

Entropy change of Silicon = m*Cp ln (Tf / T1_silicon)

= 0.2*0.712*ln ((246.8+273) / (800+273))

= -0.103 kJ/K

Entropy change of air = m_air*[Cp* ln (Tf / T1) - R ln (P2 / P1)]

But P2 / P1 = Tf / T1....since constant volume process

So, Entropy change of air = m_air*(Cp - R)* ln (Tf / T1)

= 0.3554*(1 - 0.287)*ln ((246.8+273) / (25+273))

= 0.141 kJ/K

Total entropy change = -0.103 + 0.141 = 0.038 kJ/K

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