In an experiment, 270 g of aluminum(with a specific heat of 900 J/kg·K) at 100°C

Question

In an experiment, 270 g of aluminum(with a specific heat of 900 J/kg·K) at 100°C is mixedwith 50.0 g of water (4186J/kg·K) at 20°C, with the mixture thermally isolated. (a) What is the equilibrium temperature?
1°C

(b) What is the entropy change of the aluminum?
2 J/K

(c) What is the entropy change of the water?
3 J/K

(d) What is the entropy change of the aluminum-water system?
4 J/K (a) What is the equilibrium temperature?
1°C

(b) What is the entropy change of the aluminum?
2 J/K

(c) What is the entropy change of the water?
3 J/K

(d) What is the entropy change of the aluminum-water system?
4 J/K

Explanation / Answer

heat released by aluminum = heat absorbed by water 270 g *(900 J/kg/C)*(100 C - Tfinal) = 50g *(4186 J/kg/C)*(Tfinal -20) 1.1610129*(100 - Tfinal) = (Tfinal - 20) Tfinal*(1 + 1.1610129) = 1.1610129*100 + 20 Tfinal = 62.9803228 C = 335.980323 K b) S = integral [mass* C*dT/T] = mass*C*ln(Tfinal/Tinitial) =(0.270 kg)*(900 J/kg/K)*ln(335.98 K / 373 K) S = -25.3997752 = -25.4 J/K c) S = integral [mass* C*dT/T] = mass*C*ln(Tfinal/Tinitial) =(0.050 kg)*(4186 J/kg/K)*ln(335.98 K/ 293 K) S = 28.6489812 = 28.6 J/K d) Sum both S Soverall = -25.4 + 28.6 = 3.249206 = 3.25 J/K

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