Answer :
Final answer:
The change in internal energy when 1.521 kg of water boils at 101 kPa and is replaced by 1700 L of steam is calculated using the latent heat of vaporization of water, which is 2256 kJ/kg. The total energy required for the phase change is 3430.016 kJ, which is approximately equal to the change in the internal energy of the system at constant pressure.
Explanation:
To determine the change in internal energy when 1.521 kg of water boils at 101 kPa and is replaced by 1700 L of steam, we first need to calculate the energy required to vaporize the given mass of water. The latent heat of vaporization of water is 2256 kJ/kg. Therefore, the energy needed to convert the water to steam is given by:
Qv = m × Lv
Qv = 1.521 kg × 2256 kJ/kg
Qv = 3430.016 kJ
The change in internal energy for a system can be represented by the change in heat, since the work done by the system during a phase change at constant pressure is the product of pressure and change in volume. The volume change from water to steam is substantial, as each kg of water expands to 1700 L (1.7 m3) of steam. Since internal energy is a state function, it depends only on the initial and final states and not on the path, so it is equivalent to the heat added if the process is at constant pressure. Hence, the change in internal energy (ΔU) is approximately equal to the heat supplied:
ΔU ≈ Qv
In this case, the change in internal energy when 1.521 kg of water boils and is converted to steam at 101 kPa is approximately 3430.016 kJ.
