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Answer :
Final answer:
To heat 3.50 g of ice from -10.0 °C to 27.0 °C, the energy required is calculated in three steps: warming the ice to its melting point, melting the ice, and warming the water to the target temperature, totaling 1.636 kJ.
Explanation:
Calculating this involves three steps: heating the ice to 0 °C, melting the ice, and then heating the resulting water to 27.0 °C.
To begin, we need the mass of ice in moles, molar heat capacities, and the heat of fusion. The molar mass of water is 18.015 g/mol, so 3.50 g of ice is 3.50 g / 18.015 g/mol = 0.194 moles.
- Heating the ice to 0 °C: q = mcΔT, where m is moles, c is molar heat capacity, and ΔT is the temperature change. Assuming the molar heat capacity for ice as 36.6 J/(mol·K), the energy required is q = 0.194 mol * 36.6 J/(mol·K) * 10 K = 71 J or 0.071 kJ.
- Melting the ice: The heat of fusion is needed here, q = mΔHf, with ΔHf as the heat of fusion per mole. Using 6.01 kJ/mol, q = 0.194 mol * 6.01 kJ/mol = 1.17 kJ.
- Heating water to 27.0 °C: With water's molar heat capacity at 75.4 J/(mol·K), q = 0.194 mol * 75.4 J/(mol·K) * 27 K = 395 J or 0.395 kJ.
Adding these energies gives 0.071 kJ + 1.17 kJ + 0.395 kJ = 1.636 kJ as the total energy required.
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