Final answer:
The enthalpy change (ΔH) for the melting of ice is +1440 calories, as the process involves heat absorption and is endothermic. The internal energy change (ΔE) also considers the negligible work done due to volume change and is approximately equal to +1440 calories for the conditions given.
Step-by-step explanation:
To calculate the enthalpy change (ΔH) and internal energy change (ΔE) for the melting of ice, we must consider both the heat absorbed during the phase change and the work done by the system due to volume change. It is given that 1 mole of ice absorbs 1440 calories of heat when it melts at 0°C under a constant pressure of 1 atm.
ΔH represents the total heat absorbed at constant pressure and is equal to the heat required to melt the ice, which is +1440 calories per mole as the process is endothermic. On the other hand, ΔE (change in internal energy) also accounts for the p-V work done during the melting, which can be calculated using the formula ΔE=ΔH-PΔV where P is the pressure and ΔV is the change in molar volume.
The molar volumes of ice and water are given as 0.0196 L and 0.0180 L, respectively, so the change in volume (ΔV) is 0.0180 L - 0.0196 L = -0.0016 L or -1.6 mL. Since the pressure is in atm and volume in liters, to convert calories to liter-atmospheres, we use the conversion factor 1 cal = 41.84 mL·atm. Therefore, the work done is PΔV = (1 atm) × (-1.6 mL) = -1.6 mL·atm, or when converted to calories, approximately -1.6 mL·atm × 1/41.84 mL·atm/cal = -0.038 cal.
The value is very small compared to the heat absorbed and can often be considered negligible. Hence, ΔE is approximately equal to ΔH in this context, and both are 1440 calories (positive since the system absorbs heat).