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2 July, 10:56

Consider a process in which 100 kJ of energy is transferred reversibly and isothermally as heat to a large block of copper. Calculate the change in entropy of the block if the process takes place at (i) 0 °C, (ii) 50 °C.

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  1. 2 July, 11:01
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    a) at 0°C the entropy change is ΔS=0.366 kJ/K

    b) at 50°C the entropy change is ΔS = 0.309 kJ/K

    Explanation:

    from the second law of thermodynamics

    ΔS ≥ ∫ dQ / T

    where ΔS = change in entropy, Q = heat exchanged with the surroundings

    and T = absolute temperature

    - For an isothermal process ∫ dQ / T = 1/T * ∫ dQ = Q/T

    - For an reversible process the ΔS = ∫ dQ / T

    thus

    ΔS = Q/T

    replacing values

    a) 0°C = 273 K

    ΔS = Q/T = 100 kJ / 273K = 0.366 kJ/K

    ΔS=0.366 kJ/K

    b) 50°C = 50 + 273K = 323K

    ΔS = Q/T = 100 kJ / 323K = 0.309 kJ/K

    ΔS = 0.309 kJ/K
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