Answer:
13.428° C
Explanation:
Given:
Mass of the metal piece, m = 1.9 kg
specific heat of the metal, C = 202 J/kg°C
Initial temperature of the metal, T₁ = 83° C
Final temperature of the metal, T₂ = 16° C
mass of the liquid, M = 4 kg
specific heat of the liquid, C' = 1000 J/kg°C
Initial temperature of the liquid, T₃ = 7° C
Now,
when the metal is removed
The heat lost by the metal = The heat gained by the liquid
thus,
mCΔT = MC'ΔT'
let the final temperature of the liquid be T₄
thus, we have
- [1.9 × 202 × (T₂ - T₁)] = 4 × 1000 × (T₄ - T₃) (here, negative sign depicts the heat lost by the metal)
on substituting the further values, we get
- [1.9 × 202 × (83 - 16)] = 4 × 1000 × (T₄ - 7)
or
T₄ - 7 = 6.42865
or
T₄ = 13.428° C
Hence, the final temperature of the liquid is 13.428° C