A 74.6-g ice cube floats in the Arctic Sea. The temperature and pressure of the system and surroundings are at 1 atm and 0°C. Calculate ∆Ssys, ∆Ssurr, and ∆Suniv for the melting of the ice cube. What can you conclude about the nature of the process from the value of ∆Suniv? (The molar heat of fusion of water is 6.01 kJ/mol.)
Answers
Answer: We must do calculations for the S Sys, S Surr, and S Univ. Finding the system's enthalpy change will help us do this.
Action 2 of 5
Finding the system's enthalpy change is the first step.
H is equal to 74.6 grimes of water vapor and one mole of hydrogen.
O ( ) 18.02 g H 2 O ( ) 6.01 kJ 1 mol H 2 O ( ) H sys = 74.6 g H 2 O (s) 18.02 g H 2 O 1 mol H 2 O
(s)
1 mol of H 2 O in (s).
(s)
6.01 kJ
Δ H = 24.9 kJ
H sys equals 24.9 kJ
Action 3 of 5
Next, you must determine the entropy change of the
Explanation: system: Δ S = 24.9 × 1 0 3 J 273 K
S sys = 273 K 24.9 10 3 J
Δ S = 91.2 J/K
S sys = 91.2 J/K.
As a result, S = 91.2 is the system's entropy change.
J/K = 91.2 J/K S sys
Action 4 of 5
Then, we must determine how the environment's entropy will change:
Δ S = − 24.9 × 1 0 3 J 273 K
'S surr' equals 273 K 24.9 10 3 J.
Δ S = − 91.2 J/K
S surr = 91.22 J/K
As a result, S = 91.2 is the entropy change of the environment.
J/K = S surr = 91.2 J/K
Instance 5 of 5
Finally, we must determine the universe's change in entropy:
S = S + S S univ = S sys + S surr
S = 91.2 J/K S university 91.2 J/K =91.2 J/K−91.2 J/K
Δ S = 0 J/K
J/K S univ = 0
As a result, S = 0 represents the universe's entropy change.
S univ = 0 J/K J/K. We may conclude that the melting of the ice cube at 0 o C is an equilibrium process since the entropy of the cosmos remains unaffected.
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