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### Thermal Properties of Matter NCERT Solutions Class 11 Physics - Solved Exercise Question 11.14

Question 11.14:
In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
Solution:
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 = T1T2 = 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′′ = m1 CwΔT
= (M + m′) Cw ΔT … (ii)
Heat lost by the metal = Heat gained by the water and colorimeter system
mCΔT = (M + m) Cw ΔT
200 × C × 110 = (150 + 25) × 4.186 × 13
∴ C = (175 X 4.186 X 13) / (110 X 200)  =  0.43 Jg-1K-1
If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.