3. Calculate the entropy changes of system and surroundings for the case of freezing of supercooled liquid silver at 1073 K. The melting point of silver is 1234 K and the heat of fusion is Hf = 11.2 kJ/mol. C' = 30.5J / mol / Kand C;=21.2+ 8.55×10°T +1.5×10°T*J/mol / K. HINT: Assume that the same %3D function for cp=f(T) can be used for the liquid silver even below melting point.

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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**Problem 3: Entropy Changes in Supercooled Silver**

Calculate the entropy changes of the system and surroundings for the case of freezing of supercooled liquid silver at 1073 K. The melting point of silver is 1234 K and the heat of fusion is \( H_f = 11.2 \, \text{kJ/mol} \).

For the specific heat capacities, use:
- \( C_p^l = 30.5 \, \text{J/mol/K} \) for liquid silver.
- \( C_p^s = 21.2 + 8.55 \times 10^{-3} T + 1.5 \times 10^{5} T^{-2} \, \text{J/mol/K} \) for solid silver.

**HINT:**
Assume that the same function for \( c_p = f(T) \) can be used for the liquid silver even below the melting point.
Transcribed Image Text:**Problem 3: Entropy Changes in Supercooled Silver** Calculate the entropy changes of the system and surroundings for the case of freezing of supercooled liquid silver at 1073 K. The melting point of silver is 1234 K and the heat of fusion is \( H_f = 11.2 \, \text{kJ/mol} \). For the specific heat capacities, use: - \( C_p^l = 30.5 \, \text{J/mol/K} \) for liquid silver. - \( C_p^s = 21.2 + 8.55 \times 10^{-3} T + 1.5 \times 10^{5} T^{-2} \, \text{J/mol/K} \) for solid silver. **HINT:** Assume that the same function for \( c_p = f(T) \) can be used for the liquid silver even below the melting point.
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