Question

A mixture of 1.78 kg of water and 262 g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0ºC, (a) Calculate the entropy change of the system during this process. (b) The system is then returned to the first equilibrium state, but in an irreversible way (by using a Bunsen burner, for instance). Calculate the entropy change of the system during this process. (c) Show that your answer is consistent with the second law of thermodynamics.

Answer 1

Answer:

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Explanation:

Solution:-

(a) Mass of water = 1.78 kg

Mass of ice = 262 g

So the total mass of ice and water mixture will be,

Mass of ice-water mixture = (Mass of water) + (Mass of ice)

= (1.78 kg) + (262 g) = (1.78 kg) + (262 g×10-3 kg/1 g)

= 1.78 kg + 0.262 kg = 2.04 kg

If eventually the ice and water have the same mass, then the final state will have 1.02 kg (2.04 kg/2) of each.

Thus the mass of the water that changed into ice m will be the difference of mass of water mw  and mass of final state ms.

So, m = mw - ms

To obtain mass of water that changed into ice m, substitute 1.78 kg for mass of water mw and 1.02 kg for mass of final state ms in the equation m = mw - ms,

m = mw - ms

= 1.78 kg – 1.02 kg

= 0.76 kg

The change of water at 0° C to ice at 0° C  is isothermal.

To obtain the change in entropy ΔS of the system during this process, substitute 0.76 kg for mass m, 333×103 J/kg for heat of fusion of water L and 273 K for T in the equation ΔS = -mL/T,

ΔS = -mL/T

 = -(0.76 kg) (333×103 J/kg )/(273 K)

= -927 J/K

From the above observation we conclude that, the change in entropy ΔS of the system during this process will be -927 J/K.

(b) Now the system is returned to the first equilibrium state, but in an irreversible way. Thus the change in entropy ΔS of the system during this process is equal to the negative of previous case.

So, ΔS = -(- 927 J/K)

= 927 J/K

From the above observation we conclude that, the change in entropy ΔS of the system would be 927 J/K.

(c) In accordance to second law of thermodynamics, entropy change ΔS is always  ≥  zero.

The total change in entropy will be,

ΔS = (-927 J/K) + (927 J/K) = 0

From the above observation we conclude that, our answer is consistent with the second law of thermodynamics.

Answer 2

Answer:

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Explanation:

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