Given, mass of water, m=lg Volume of 1 g of water =1 cm3=10−6 m3 Volume of 1 g of steam =1671 cm3=1671×10−6 m3 Pressure, p=1×105 Pa Latent heat of vaporization of water, L=2256 J/g Change in volume, ΔV=(1671−1)×10−6 m3 =1670×10−6 m3 Heat supplied, ΔQ=mL=1×2256=2256 J As the steam expands, so the work done in expansion is ΔW=pΔV=1×105×1670×10−6 [from Eq. (i)] =167 J….(iii) According to first law of thermodynamics, ΔQ=ΔU+ΔW⇒ΔU=ΔQ−ΔW=2256−167=2089 J [from Eq. (ii) and (iii)]