Answer:
the final steady temperature of the mixture is 46.67 ⁰C
Explanation:
Given;
mass of first water,m₁ = 6 g = 0.006 kg
initial temperature, t₁ = 50 ⁰C
mass of the second water, m₂ = 3.0 g = 0.003 kg
initial temperature of the second water, t₂ = 40 ⁰C
specific heat capacity of water, C = 4,200 J/kg.K
Let the final temperature of the mixture = T
Based on Law of conservation of energy, the temperature of the final mixture is calculated as follows;
m₁c(t₁ - T) = m₂c(T - t₂)
0.006 x 4200 x (50 - T) = 0.003 x 4200 x (T - 40)
1,260 - 25.2T = 12.6T - 504
1,260 + 504 = 12.6T + 25.2T
1764 = 37.8T
T = 1764/37.8
T = 46.67 ⁰C
Therefore, the final steady temperature of the mixture is 46.67 ⁰C
