1 g of water (1 cm³) becomes 1671 cm³ of steam at $1.013\times10^5$ Pa. The external work done is about:

1. Work done against a constant external pressure is $W=P\Delta V$. 2. The volume change is $\Delta V=1671-1=1670$ cm³. 3. Convert to SI: $1670\text{ cm}^3=1670\times10^{-6}=1.670\times10^{-3}$ m³. 4. So $W=1.013\times10^5\times1.670\times10^{-3}$. 5. That gives about $169$ J. Forgetting the cubic-centimetre conversion shifts the answer by a factor of 1000. _Source: NECTA ACSEE 2023 Physics 131/1, Question 6: Heat (First law of Thermodynamics)_