Mass of steam, ms=1kg=1000g=103 Temperature of steam, T1=150∘C Latent heat of steam, Ls=540calg−1 c‌‌=1calg−1‌∘C−1 Heat lost by steam, Q′=mLf+msc∆t
=103×540+103×1×(150−90)
=54×104+6×104 =104(54+6)=60×104cal Heat gained by 20L water, Q′′=mw×c×∆T Mass of 20L water, mw=20kg=20×103g ∴Q"=20×103×1×∆T By the principle calorimetry, Heat lost = Heat gained 60×104=20×103×∆T ⇒‌‌∆T=‌