According to the First Law of Thermodynamics: $\Delta Q = \Delta U + \Delta W$ Where: $\Delta Q$ is the heat absorbed, $\Delta U$ is the change in internal energy, $\Delta W$ is the work done by the system.

Step 1: Convert Heat into Joules Given $\Delta Q = 54 \text{ cal}$. Using the conversion $1 \text{ cal} = 4.18 \text{ J}$: $\Delta Q = 54 \times 4.18 \text{ J} = 225.72 \text{ J}$

Step 2: Calculate Work Done ($\Delta W$) The process occurs at constant pressure (isobaric). $\Delta W = P \Delta V$ Given $P = 1.013 \times 10^5 \text{ N m}^{-2}$ Change in volume, $\Delta V = V_{steam} - V_{water}$. Since the volume of liquid water ($0.1 \text{ cc}$) is negligible compared to steam ($167.1 \text{ cc}$), we take $\Delta V \approx 167.1 \text{ cc}$. $\Delta V = 167.1 \times 10^{-6} \text{ m}^3$ $\Delta W = (1.013 \times 10^5) \times (167.1 \times 10^{-6}) \text{ J}$ $\Delta W = 1.013 \times 16.71 \text{ J} \approx 16.93 \text{ J}$

Step 3: Calculate Change in Internal Energy ($\Delta U$) $\Delta U = \Delta Q - \Delta W$ $\Delta U = 225.72 \text{ J} - 16.93 \text{ J} = 208.79 \text{ J}$

Rounding to one decimal place, $\Delta U \approx 208.7 \text{ J}$.