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Grade: 11
        
A block of aluminium originally at 80ºC is placed into an insulated container of water originally at 25ºC. After a while the system reaches an equilibrium temperature of  31ºC.                                                                  
(A) ∆Saluminum > 0.
(B) S­aluminum = 0.
(C) ∆Saluminum
(b) during this process
 (A) ∆Swater > 0.
 (B) ∆Swater = 0.               
 (C) ∆Swater
 (C) During this process
 (A) | ∆Swater | > | ∆Saluminum|.
 (B) |∆Swater| = |∆Saluminum|.
 (C) |∆Swater| aluminum|.
3 years ago

Answers : (1)

Aditi Chauhan
askIITians Faculty
396 Points
							a)

The correct option is (C) ΔSaluminum<0.
Entropy change (ΔS) for a reversible process is defined as,
ΔS = ΔQ/T
Where, ΔQ is the change in heat energy that is transferred into or out of the closed system at constant temperature T.
Whenever an irreversible process takes place, certain amount of energy which could have been utilized for doing useful work changes to a form in which it becomes unavailable. Entropy is a measure of disorder. Thus with disorder, entropy increases. The initial temperature of the aluminum block is 80̊C. But after some time the system reaches an equilibrium temperature of 31̊C. Therefore the aluminum block will colder than its original state. This signifies that, the change in entropy of the block of aluminum will be less than zero (ΔSaluminum<0). From the above observation we conclude that option (C) is correct.
(b)
The correct option is (A) ΔSwater>0.
Entropy change (ΔS) for a reversible process is defined as,
ΔS = ΔQ/T
Where, ΔQ is the change in heat energy that is transferred into or out of the closed system at constant temperature T.
Whenever an irreversible process takes place, certain amount of energy which could have been utilized for doing useful work changes to a form in which it becomes unavailable. Entropy is a measure of disorder. Thus with disorder, entropy increases. The initial temperature of water is 25̊ C. But after some time the system reaches an equilibrium temperature of 31̊C. Therefore the water will be much hotter than its original state. This signifies that, the change in entropy of the water will be greater than zero (ΔSwater>0). From the above observation we conclude that option (A) is correct.
(c)
The correct option is (A) |ΔSwater| > | ΔSaluminum|
In accordance to second law of thermodynamics, when changes occur within a closed system its entropy increases for an irreversible process. To satisfy the second law of thermodynamics in the system, the net change in entropy of water must be greater than the net change in entropy of aluminum (|ΔSwater| > | ΔSaluminum|). From the above observation we conclude that, option (A) is correct.
3 years ago
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