An insulated Thermos contains 114 g of water at 94°C. You put in a 13.5 g ice cu

Question

An insulated Thermos contains 114 g of water at 94°C. You put in a 13.5 g ice cube at 0°C to form a systemof ice + original water. (a) What is the equilibrium temperature of the system?
°C

(b) What are the entropy changes of the water that was originallythe ice cube as it melts and as it warms to the equilibriumtemperature?
melting process J/K warming process J/K
(c) What is the entropy change of the original water as it cools tothe equilibrium temperature?
J/K

(d) What is the net entropy change of the ice + original watersystem as it reaches the equilibrium temperature?
J/K
(a) What is the equilibrium temperature of the system?
°C

(b) What are the entropy changes of the water that was originallythe ice cube as it melts and as it warms to the equilibriumtemperature?
melting process J/K warming process J/K
(c) What is the entropy change of the original water as it cools tothe equilibrium temperature?
J/K

(d) What is the net entropy change of the ice + original watersystem as it reaches the equilibrium temperature?
J/K
melting process J/K warming process J/K

Explanation / Answer

the temperature equilibrium of the system is T 114*10-3*(94-T)*4200=3.34*105*13.5*10-3+ 4200*13.5*10-3*(T-0) => T=75.60 (b) in the melting process: Temperature is constant otherwise: dE=dQ/T so entropychanges=3.34*105*13.5*10-3/273=16.5 J/K in the warming process dQ=4200*mass*dT=k*dT =>dE=kdT/T=>E=k*ln(T2/T1)=4200*13.5*10-3*ln(348.6/273)=13.86J/K (c) similar to part b: E for originalwater=4200*114*10-3*ln(348.6/367)=-24.6 J/K (d) final E=16.5 + 13.86 -24.6=5.76 J/K

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