**Solution**

When 1 kg of lead is melted, 2.45 × 10^{4} J of heat is added to it. Therefore, when 3 × 10^{2} g = 3 × 10^{-1} kg of lead is melted 2.45 × 10^{4} × 3 × 10^{-1} = 7.35 × 10^{3} J of heat ΔQ is added to it. Similarly, the temperature is 327° C = 327 + 273 = 600 K. So, apply formula ΔS = ΔQ/T,