After 200.00 gm of gold ( C p = 0.03 cal / K -gm) at 12 7 ∘ C is dropped into 4.00 gm of water at 2 7 ∘ C ( C p = 1.00 cal / K − gm ) , the system is allowed to reach equilibrium in an adiabatic c

Question

After 200.00 gm of gold ( C p ​ = 0.03 cal / K -gm) at 12 7 ∘ C is dropped into 4.00 gm of water at 2 7 ∘ C ( C p ​ = 1.00 cal / K − gm ) , the system is allowed to reach equilibrium in an adiabatic container. Which of the following is/are correct information(s)? ( ln 2 = 0.70 , ln 3 = 1.10 , ln 5 = 1.60 )

TestHub · Accepted Answer

Final Temperature ( T f ​ ): From heat lost = heat gained, 200 × 0.03 × ( 127 − T f ​ ) = 4 × 1 × ( T f ​ − 27 ) ⟹ T f ​ = 8 7 ∘ C ( or360K ) . Log Values: Using given values, ln 10 = ln 2 + ln 5 = 2.30 . Entropy of Gold: Δ S A u ​ = 6 ln ( 400 360 ​ ) = 6 ( 2 ln 3 − ln 10 ) = 6 ( 2.20 − 2.30 ) = − 0.6cal/K . Entropy of Water: Δ S w a t er ​ = 4 ln ( 300 360 ​ ) = 4 ( ln 3 + 2 ln 2 − ln 10 ) = 4 ( 2.50 − 2.30 ) = + 0.8cal/K . Total Entropy: Δ S sys ​ = Δ S A u ​ + Δ S w a t er ​ = − 0.6 + 0.8 = + 0.2cal/K .

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