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17 September, 17:33

A 45-kg iron block initially at 280°C is quenched in an insulated tank that contains 100 kg of water at 18°C. Assuming the water that vaporizes during the process condenses back in the tank, determine the total entropy change during this process. The specific heat of water at 25°C is cp = 4.18 kJ/kg·K. The specific heat of iron at room temperature is cp = 0.45 kJ/kg·K

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  1. 17 September, 21:04
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    Answer: - 3.46kJ/K

    Explanation:

    From the question above, we have:

    The mass of the block (m) = 45kg

    The initial temperature of the block (T1) = 280∘C

    The weight of the water (mw) = 100kg

    The temperature of water (Tw) = 18∘C

    Recall the energy balance equation,

    ΔUI = - ΔUw

    In this case ΔUI is the internal energy of the iron, while ΔUw is the internal energy of water.

    [mcp (T2 - T1) ]I = - [mcp (T2 - T1) ]w

    Here cp is the specific heat at constant pressure.

    The specific heat of iron is (cp) I = 0.45kJ/kg⋅K, and the specific heat of water is (cp) w = 4.18kJ/kg⋅K.

    Now, we substitute the values in above equation,

    [45 * 0.45 (T2 - 280) ]I = - [100 * 4.18 (T2 - 18) ]w

    [20.25 (T2 - 280) ] = - [418 (T2 - 18) ]

    20.25T2 - 5,670 = - [418T2 - 7,524]

    20.25T2 - 5,670 = - 418T2 + 7,524

    20.25T2 + 418T2 = 7,524 + 5,670

    438.35T2 = 13,194

    T2 = 30.1K

    Recall, the expression to calculate the total entropy change is given as:

    ΔStotal = ΔSI + ΔSw

    ΔStotal = [mcpln (T2/T1) ]I + [mcpln (T2/T1) ]w

    Now, we substitute the values in above equation,

    ΔStotal = [45 * 0.45ln (297.6/553) ]I + [100 * 4.18ln (297.6/291) ]w

    ΔStotal = - 12.55 + 9.09

    ΔStotal = - 3.46kJ/K

    Thus the total entropy change is - 3.46kJ/K.
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