Find the increase in entropy of 1.00 kg of ice originally at 0 degrees Centigrade, that is melted to form water

Here's how to calculate the increase in entropy: To find the increase in entropy ($\Delta S$) during a phase change at constant temperature, we use the formula: $$\Delta S = \frac{Q}{T}$$ where $Q$ is the heat absorbed and $T$ is the absolute temperature in Kelvin. For melting, the heat absorbed ($Q$) is given by: $$Q = mL_f$$ where $m$ is the mass and $L_f$ is the latent heat of fusion. Given values: • Mass of ice, $m = 1.00 \text{ kg}$ • Temperature, $T_C = 0^\circ \text{C}$ • Latent heat of fusion of water, $L_f = 334 \times 10^3 \text{ J/kg}$ (standard value) Step 1: Convert the temperature from Celsius to Kelvin. $$T_K = T_C + 273.15$$ $$T_K = 0 + 273.15 = 273.15 \text{ K}$$ Step 2: Calculate the heat absorbed ($Q$) to melt the ice. $$Q = mL_f$$ $$Q = (1.00 \text{ kg}) \times (334 \times 10^3 \text{ J/kg})$$ $$Q = 334 \times 10^3 \text{ J}$$ Step 3: Calculate the increase in entropy ($\Delta S$). $$\Delta S = \frac{Q}{T_K}$$ $$\Delta S = \frac{334 \times 10^3 \text{ J}}{273.15 \text{ K}}$$ $$\Delta S \approx 1222.84 \text{ J/K}$$ $$\boxed{\Delta S = 1223 \text{ J/K}}$$ What's next? Send 'em!