As the surrounding is identical, vessel is identical time taken to cool both water and liquid (from 30∘C to 25∘C ) is same 2 minutes, therefore (dtdQ)water =(dtdQ)liquid or, t(mWCW+W)ΔT=t(mℓCℓ+W)ΔT (W= water equivalent of the vessel ) or, mwCw=mℓCℓ ∴ Specific heat of liquid, Cℓ=mℓmWCW =10050×1=0.5kcal/kg