Answer :
Final answer:
Assuming the provided specific energy replacement value of 1700 J/kg for boiling water, the total change in internal energy for 1.682 kg of water is 2860.4 J.
Explanation:
To determine the change in internal energy when 1.682 kg of water boils at 101 kPa, we can use the latent heat of vaporization of water. The latent heat of vaporization is the energy required to turn liquid water into water vapor without changing its temperature. For water, this value is 2256 kJ/kg at 100°C and 1 atm pressure. Using this value, the amount of energy needed for vaporization is given by the equation Qv = m * Lv.
Qv = 1.682 kg * 2256 kJ/kg = 3796.832 kJ or 3796832 J. However, the question states that each kg of boiling water is replaced by 1700 J (which seems an incorrect statement, as it typically would be much higher), which we will take as the given data for this calculation, contrary to the standard latent heat of vaporization for water. Thus, the total energy replaced when the water vaporizes would be:
Qv' = 1.682 kg * 1700 J/kg = 2860.4 J.
Therefore, assuming each kilogram of water consumes 1700 J when boiled away by the provided information, the total change in internal energy for 1.682 kg of water would be 2860.4 J.
