Answer :
Final answer:
The change in internal energy when boiling 1.217 kg of water at 101 kPa is determined by the energy required for the phase change and the work done by the steam against atmospheric pressure. This is calculated using the latent heat of vaporization and the first law of thermodynamics.
Explanation:
The change in internal energy for boiling water can be calculated using the concept of latent heat. When 1.217 kg of water is boiled and converted to steam at 101 kPa, the energy required for the phase change is the product of the mass and the latent heat of vaporization. The latent heat of vaporization of water at 100°C is 2256 kJ/kg. Thus, the total energy (Qv) needed to vaporize the water is Qv = m * Lv = 1.217 kg * 2256 kJ/kg, resulting in a change in internal energy represented by this value.
Additionally, we must consider the work done by the expanding steam against atmospheric pressure. The work done (W) by an expanding gas can be calculated as W = PΔV, where P is the pressure and ΔV is the change in volume. In this case, the pressure is given as 101 kPa, and if each kg of boiling water is replaced by 1700 L of steam, we can convert this volume to cubic meters (1 L = 0.001 m³) and multiply by the mass of the water to find the total change in volume. The work done is then subtracted from the thermal energy (Q) to find the change in internal energy (ΔEint) from the first law of thermodynamics: ΔEint = Q - W.
