Answer :
Final answer:
The change in internal energy when 1.682 kg of boiling water is converted into steam at 101 kPa is found by multiplying the mass of water by the enthalpy of vaporization, resulting in a total of 3794.272 kJ.
Explanation:
The change in internal energy when 1.682 kg of water boils at 101 kPa can be found by considering the energy required for the phase change from water to steam. When boiling water turns into steam, it absorbs a significant amount of heat, known as the enthalpy of vaporization, without changing temperature. In this case, each kg of boiling water is replaced by 1700 L of steam (water vapor). According to the information provided, it takes 2256 kJ to vaporize 1 kg of liquid water at the normal boiling point to steam. Therefore, to find the total change in internal energy for 1.682 kg, the enthalpy of vaporization would be multiplied by the mass of the water being converted to steam.
Q = m × ∆Hvap = 1.682 kg × 2256 kJ/kg = 3794.272 kJ
This calculation determines the energy required to vaporize the given amount of water which directly relates to the change in internal energy during the phase change at constant pressure.
