Final answer:
When 1g of ice is added to 4g of water at 10°C, only a fraction of the ice will melt before reaching a final temperature of 0°C with some ice remaining. The entropy of the system increases during this process.
Step-by-step explanation:
Assuming that all the heat from the water at 10°C is used to melt the ice at 0°C, the two will likely reach a thermal equilibrium. The heat required to melt ice (enthalpy of fusion) is 333.55 kJ/kg. The 4g of water at 10°C has a specific heat capacity of 4.184 J/(g·°C). The amount of heat released as the water cools down to 0°C is q = mcΔT = 4g * 4.184 J/(g·°C) * 10°C = 167.36 J.
The heat required to melt 1g of ice is q = mLf = 1g * 333.55 kJ/kg = 333.55 J (since we need to convert 1g to kg by dividing by 1000). The amount of ice that can be melted by the cooling water is found by the amount of heat the water can give up, which is 167.36 J. Since this is not enough to melt the 1g ice completely (requires 333.55 J), only a fraction of the ice will melt, and the final temperature will be 0°C, with some ice remaining. As heat is distributed evenly in the system, and ice is at 0°C, no additional cooling or heating takes place. Therefore, the entropy change, ΔS, considering entropy increases upon melting, would be greater than 0.
Based on the process and available heat, the best-fitting answer would be b) Ice melts, final T = 0°C, some ice left, ΔS > 0.