We have, Mass of the metal, m = 0.20 kg = 200 gInitial temperature of the metal, T1 = 150°CFinal temperature of the metal, T2 = 40°CCalorimeter has water equivalent of mass, m’ = 0.025 kg = 25 gVolume of water, V = 150 cm3 Mass (M) of water at temperature T = 27°C is,                     150 × 1 = 150 g Fall in the temperature of the metal,          ΔT = T1 – T2                = 150 – 40              = 110°C Specific heat of water, Cw = 4.186 J/g/°K Specific heat of the metal = C Heat lost by the metal, θ = mCΔT                       … (i) Rise in the temperature of the water and the calorimeter system is,              ΔT′’ = 40 – 27 = 13°C Heat gained by the water and the calorimeter system is,              Δθ′′ = m1 CwΔT’                    = (M + m′) Cw ΔT’                        … (ii)Heat lost by the metal = Heat gained by the water and colorimeter system.That is,                  mCΔT = (M + m’) Cw ΔT’      200 × C × 110 = (150 + 25) × 4.186 × 13If some heat is lost to the surroundings, then the value of C will be less than the actual value.