We need to calculate the work done and the increase in external energy. The external work can be calculated as:
[tex]W=P\Delta V[/tex]where P is the pressure and delta V denotes the change in volume.
The increase en external energy is given by:
[tex]\Delta E=Q-W[/tex]where W is the work done by external forces and Q is the heat transfer.
Before we begin we convert the units to the correct units:
[tex]\begin{gathered} m=0.001kg \\ V_1=1\times10^{-6}m \\ V_2=1671\times10^{-6}m \\ P=101325\text{ Pa} \end{gathered}[/tex]a)
Plugging the values we know we have that:
[tex]\begin{gathered} W=101325(1671\times10^{-6}-1\times10^{-6}) \\ W=169.21 \end{gathered}[/tex]Therefore, the work done is 169.21 J
b)
The heat transfer is given by:
[tex]Q=mL[/tex]then we have:
[tex]\begin{gathered} Q=(0.001)(2260\times10^3) \\ Q=2260 \end{gathered}[/tex]Once we know this we have that the change in external energy is:
[tex]\begin{gathered} \Delta E=2260-169.21 \\ \Delta E=2090.79 \end{gathered}[/tex]Therefore, the increase in external energy is 2090.79 J