Answer :
The change in internal energy when 1.416 kg of water boils at 101 kPa is 3674.304 kJ. It can be calculated using the equation:ΔU = m × (ΔHv + ΔHf).
where ΔU is the change in internal energy, m is the mass of water, ΔHv is the enthalpy of vaporization, and ΔHf is the enthalpy of fusion.
First, calculate the enthalpy of vaporization (ΔHv). The enthalpy of vaporization of water is 2260 kJ/kg.
ΔHv = 2260 kJ/kg
Next, calculate the enthalpy of fusion (ΔHf). The enthalpy of fusion of water is 334 kJ/kg.
ΔHf = 334 kJ/kg
Now, calculate the change in internal energy:
ΔU = 1.416 kg × (2260 kJ/kg + 334 kJ/kg)
ΔU = 1.416 kg × 2594 kJ/kg
ΔU = 3674.304 kJ
Therefore, the change in internal energy when 1.416 kg of water boils at 101 kPa is 3674.304 kJ.
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