In a specific heat experiment 208.0 grams of aluminum (C=913.0 J/kg.K) at 100.0

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First step: use conservation of energy to calculate the equilibrium temperature: mw*cw*(Teq - Tw1) = mal*cal*(Tal1 - Teq) Solve for Teq: Teq = (mw*cw*Tw1 + mal*cal*Tal1)/(mw*cw + mal*cal) Entropy change of the water: dSw = dQw/T = mw*cw*dT/T Integrate from T=Tw1 to T=Teq: ?Sw = mw*cw*ln(Teq/Tw1) Entropy change for aluminum: dSal = dQal/T = mal*cal*dT/T Integrate from T=Tal1 to T=Teq: ?Sal = mal*cal*ln(Teq/Tal1) I know that Tal1 > Teq, thus, to make it more intuitive, simplify with properties of logarithms: ?Sal = -mal*cal*ln(Tal1/Teq) Total change in entropy of system: ?Snet = mw*cw*ln(Teq/Tw1) - mal*cal*ln(Tal1/Teq) No point in substitute in equilibrium temperature equation (it just makes it messy), just keep it in mind when computing. Teq = (mw*cw*Tw1 + mal*cal*Tal1)/(mw*cw + mal*cal) Data: mw:= 0.058 g; mal:=0.208 g; Tal1:=373.15 K; Tw1:=293.15 K; cal:=913 J/kg-K; cw:=4184 J/kg-K; Results: = 0.058*4184*293.15 + 0.208*913*373.15 /(0.058*4184+0.208*913) Teq = 284 Kelvin = 0.058*4184 * ln (284/293.15) - 0.208*913* ln(373.15/284) ?Snet = 0.5954 Joules/Kelvin

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