How much energy (in kJ) is required to convert 70.6 g of ice at 0.00°C to liquid water at 75.0°C? (Given: specific heat of ice = 2.09 J/g·°C, specific heat of water = 4.18 J/g·°C, enthalpy of fusion for ice = 334 J/g)

• A. 38.5 kJ
• B. 52.8 kJ
• C. 46.2 kJ
• D. 41.7 kJ

Answer: (D)

Verified step by step guidance

1. Identify the three stages involved in converting ice at 0.00°C to liquid water at 75.0°C: (1) warming the ice from 0.00°C to 0.00°C (which is zero in this case since it starts at melting point), (2) melting the ice at 0.00°C to liquid water at 0.00°C, and (3) heating the liquid water from 0.00°C to 75.0°C.
2. Calculate the energy required to melt the ice using the enthalpy of fusion: use the formula \(q_{fusion} = m \times \Delta H_{fusion}\), where \(m\) is the mass of the ice and \(\Delta H_{fusion}\) is the enthalpy of fusion (334 J/g).
3. Calculate the energy required to heat the liquid water from 0.00°C to 75.0°C using the specific heat capacity of water: use the formula \(q_{water} = m \times C_{water} \times \Delta T\), where \(C_{water} = 4.18\) J/g·°C and \(\Delta T = 75.0 - 0.00\) °C.
4. Since the ice is initially at 0.00°C, no energy is needed to warm the ice before melting, so you can skip the step involving the specific heat of ice in this particular problem.
5. Add the energies calculated for melting the ice and heating the water to find the total energy required: \(q_{total} = q_{fusion} + q_{water}\). Convert the total energy from joules to kilojoules by dividing by 1000.