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Asked by Deepani Bulkquestions Last Modified
Answer 
Harshit Bhardwaj
Mass of the metal, m = 0.20 kg = 200 g
Initial temperature of the metal, T1 = 150°C
Final temperature of the metal, T2 = 40°C
Calorimeter has water equivalent of mass, m’ = 0.025 kg = 25 g
Volume of water, V = 150 cm3
Mass (M) of water at temperature T = 27°C:
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 calorimeter system:
ΔT′’ = 40 – 27 = 13°C
Heat gained by the water and calorimeter system:
Δθ′′ = m1CwΔT’
= (M + m′) CwΔT’ … (ii)
Heat lost by the metal = Heat gained by the water and colorimeter system
mCΔT = (M + m’) CwΔT’
200 × C × 110 = (150 + 25) × 4.186 × 13
If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.
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