The specific heat of ice is 2.10 kJ/kg °C, the heat of fusion for ice at 0 °C is 333.7 kJ/kg, the specific heat of water 4.186 kJ/kg °C, and the heat of vaporization of water at 100 °C is 2,256 kJ/kg. What is the final equilibrium temperature when 5.00 grams of ice at −15.0 °C is mixed with 40.0 grams of water at 75.0 °C?

To solve this problem, we need to consider the energy transfers that occur when the ice and water are mixed. The ice will absorb heat from the water until it reaches 0 °C, then it will melt, and finally it will heat up to the final equilibrium temperature. The water will lose heat until it reaches the final equilibrium temperature. The total heat gained by the ice must equal the total heat lost by the water.

1. Calculate the heat absorbed by the ice to reach 0 °C:

   Q1 = mass_ice * specific_heat_ice * change_in_temperature Q1 = 5.00 g * 2.10 kJ/kg°C * 15.0 °C = 157.5 J

2. Calculate the heat absorbed by the ice to melt:

   Q2 = mass_ice * heat_of_fusion Q2 = 5.00 g * 333.7 kJ/kg = 1668.5 J

3. Calculate the heat lost by the water to reach 0 °C:

   Q3 = mass_water * specific_heat_water * change_in_temperature Q3 = 40.0 g * 4.186 kJ/kg°C * 75.0 °C = 12555 J

4. If Q3 > Q1 + Q2, then the final equilibrium temperature is 0 °C. If Q3 < Q1 + Q2, then some of the ice will not melt, and the final equilibrium temperature is still 0 °C. If Q3 = Q1 + Q2, then all the ice melts and the final equilibrium temperature is above 0 °C. We need to calculate the heat absorbed by the melted ice to reach this temperature:

   Q4 = mass_melted_ice * specific_heat_water * change_in_temperature

   We can solve for the change in temperature:

   change_in_temperature = (Q3 - Q1 - Q2) / (mass_melted_ice * specific_heat_water)

   If all the ice melts, then mass_melted_ice = mass_ice.

5. The final equilibrium temperature is the initial temperature of the ice plus the change in temperature.