q=ms∆t 0∘C lce (s)⟶0∘C water (I) q1=94×80=7520‌cal 0∘C water (I)⟶100∘C water (I) q2=94×1×100=9400‌cal 100∘C (water) 100∘C water vapours q3=94×540=50760‌cal Total (q)=67680‌cal =67.680‌kcal ∵‌‌94‌kcal=12‌gcoal ∴‌‌67.68‌kcal=‌