Answer:
80 °C
Step-by-step explanation:
The heat transfer parameters for the water and copper container are;
Mass of the copper container, m₁ = 84 g
Mass of the water in the container, m₂ = 84 g
Initial temperature of the water in the container, T₂ = 20°C
Mass of the hot water added, m₃ = 46 g
Initial temperature of the hot water, T₃ = 200°C
Specific heat capacity of water, c₂ = 4,200 J·kg⁻¹·°C⁻¹
Specific heat capacity of copper, c₁ = 400 J·kg⁻¹·C⁻¹
The formula for the specific heat, ΔQ = m·c·ΔT
The heat lost by the hot water = The heat gained by the container the and the cold water
The formula for the specific heat of the mixture is presented as follows;
m₃ × c₃ × (T₃ - T) = m₁ × c₁ × (T - T₁) + m₂ × c₂ × (T - T₂)
Where T represents the final temperature of the water
Therefore, by plugging in the values, we get;
46 × 4200 × (200 - T) = 84 × 400 × (T - 20) + 84 × 4200 × (T - 20)
38640000 - 193200·T = 386400·T - 7728000
38640000 + 7728000 = 46368000 = 386400·T + 193200·T = 579,600·T
∴ T = 46368000/579,600 = 80
The final temperature of the water, T = 80°C