Answer :
Final answer:
The enthalpy change for converting 80 grams of ice into water at 0 degrees Celsius and 1 atm is calculated using the latent heat of fusion, resulting in a value of 26.72 kJ.
Explanation:
The student is asking about the enthalpy change when 80 grams of ice is converted into water at 0 degrees Celsius and 1 atm pressure. To calculate the enthalpy change, the latent heat of fusion for water, which is 334 kJ/kg, is used. For 80 grams (0.08 kg) of ice, the heat transfer necessary for melting can be calculated using the formula Q = mLf, where m is the mass of ice and Lf is the latent heat of fusion. Therefore, Q = (0.08 kg)(334 kJ/kg) = 26.72 kJ.
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