Final answer:
To make ice cubes at -8.00 °C, the energy that must be removed from the water is approximately 6736.6 J.
Step-by-step explanation:
To calculate the energy that must be removed from the water to make ice cubes at -8.00 °C, we need to consider the heat absorbed in three steps:
Heating the ice from -8.00 °C to 0.00 °C: Q₁ = mass of ice × specific heat of ice × temperature change = 75.0 g × 2.062 J/(g°C) × 8.00 °C = 9.85 J.
Melting the ice at 0.00 °C: Q₂ = mass of ice × enthalpy of fusion of ice = 75.0 g × 6.01 kJ/mol = 450.75 J.
Cooling the water from 20.0 °C to 0.00 °C: Q₃ = mass of water × specific heat of water × temperature change = 75.0 g × 4.184 J/(g°C) × 20.0 °C = 6276 J.
The total energy to be removed is the sum of the three heats: Q = Q₁ + Q₂ + Q₃ = 9.85 J + 450.75 J + 6276 J = 6736.6 J (rounded to four significant figures).