100 grams of molten lead (600°C) is used to make musket balls. If the lead shot is allowed to cool to room temperature (21°C), what is the change in entropy (in J/K) of the lead? (For the specific heat of molten and solid lead use 1.29 J/g⋅°C; the latent heat of fusion and the melting point of lead are 2.45 × 10⁴ J/kg and 327°C.)

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The correct answer is A.The change in entropy is given by the equation:ΔS = (Q/T) - (Q/T)where Q is the heat transferred and T is the temperature.In this case, the heat transferred is the heat of fusion and the heat of cooling. The heat of fusion is the heat required to melt the lead, and the heat of cooling is the heat released when the lead cools to room temperature.The heat of fusion is given by the equation:Q = mLwhere m is the mass of the lead and L is the latent heat of fusion.The heat of cooling is given by the equation:Q = mcΔTwhere c is the specific heat of the lead and ΔT is the change in temperature.Substituting these equations into the equation for the change in entropy, we get:ΔS = (mL/T) - (mcΔT/T)ΔS = (100 g * 2.45 × 10⁴ J/kg) / (600°C + 273 K) - (100 g * 1.29 J/g⋅°C * (600°C - 21°C)) / (21°C + 273 K)ΔS = -145 J/K

100 Grams of Molten Lead (600°C) Is Used to Make