Question

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 the specific heat of the metal?

Answer 1

Mass of the metal, m = 0.20 kg = 200 g 
The initial temperature of the metal, T₁ = 150°C 
The final temperature of the metal, T₂ = 40°C 
The calorimeter has water equivalent of mass, m’ = 0.025 kg = 25 g 
The 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, C_w = 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 C_wΔT’ 
= (M + m′) C_w ΔT’ ...… (ii)
Heat lost by the metal = Heat gained by the water and colorimeter system
mCΔT = (M + m’) C_w ΔT’
200 × C × 110 = (150 + 25) × 4.186 × 13
∴ C = \(\frac{175 \times 4.186 \times 13}{110 \times 200}\) = 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.