Answer :
Final answer:
The change in internal energy when 1.726 kg of water boils and becomes steam at 101 kPa is calculated using the latent heat of vaporization. The calculation shows it is approximately 3894.176 kJ, which is closest to option A, acknowledging that the exact value is not an option provided.
Explanation:
To determine the change in internal energy when 1.726 kg of water boils at 101 kPa and is turned into steam, we can use the latent heat of vaporization for water and the formula Qv = m×Lv, where Qv is the heat required for vaporization, m is the mass of the water, and Lv is the latent heat of vaporization.
Using the latent heat of vaporization of water at 2256 kJ/kg, the calculation is as follows:
Qv = 1.726 kg × 2256 kJ/kg = 3894.176 kJ
To consider the work done by the water when expanding against the atmosphere, we use W = P×ΔV where P is the pressure and ΔV is the change in volume. However, in this case, the volume change multiplied by the pressure is negligible compared to the heat of vaporization, so the dominant term in the change in internal energy is the heat added for the phase change.
Therefore, the change in internal energy is most closely represented by option A. 1784.4 kJ, if we consider that this value is the closest to the calculated energy required for vaporization, acknowledging that the options given do not exactly match the calculated value.
