1.00 kg of liquid water at 100^{circ} C unde | vedclass

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(A) According to the first law of thermodynamics, $\Delta Q = \Delta U + W$. Therefore, $\Delta U = \Delta Q - W$. The heat supplied for the phase cha

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$1800$

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(A) According to the first law of thermodynamics, $\Delta Q = \Delta U + W$. Therefore, $\Delta U = \Delta Q - W$. The heat supplied for the phase change of water is $\Delta Q = m L = 1 \ kg 	imes 2000 \ kJ/kg = 2000 \ kJ$. The change in volume during the phase change is $\Delta V = V_{final} - V_{initial} = 2.001 \ m^3 - 0.001 \ m^3 = 2 \ m^3$. The work done by the steam during expansion is $W = P \Delta V = (1.00 	imes 10^5 \ Pa) 	imes (2 \ m^3) = 2 	imes 10^5 \ J = 200 \ kJ$. Substituting these values into the energy equation: $\Delta U = 2000 \ kJ - 200 \ kJ = 1800 \ kJ$.

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