The values are given in the above question, ΔQ=54cal=54×4.18joule=225.72joule
ΔW=P[Vsteam−Vwater][Forwater0.1gram=0.1cc]
ΔW=1.013×105[167.1×10−6−0.1×10−6]joule
⇒ΔW=1.013×167×10−1=16.917joule
. By first law of thermodynamics, the change in internal energy,
⇒ΔU=ΔQ−ΔW=225.72−16.917=208.8J