8. A copper container of 84g mass contains 84g of water at 20°C. 46g of water at 200°C is mixed with water in the copper ontainer. What is the final temperature of the water? Specific heat capacity of water = 4200 J kg-1 °C-1, Specific heat capacity of copper = 400 J kg-1 °C-1 ​

80 °C

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

i wanna say it’s between b and d, leaning more towards b.