A 300 g sample of copper is heated to 285°C and then placed into 1000 g of water at 26.5°C period what was the final temperature of the water?

33.4 °C

Explanation:

The final temperature of water will be the equilibrium temperature. This can be obtained as follow:

Mass of copper (M꜀) = 300 g

Initial temperature of copper (T꜀) = 285 °C

Specific heat capacity of copper (C꜀) = 0.385 J/gºC

Mass of water (Mᵥᵥ) = 1000 g

Initial temperature of water (Tᵥᵥ) = 26.5 °C

Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Equilibrium temperature (Tₑ) =?

Heat lost by copper = Heat gained by water

M꜀C꜀(T꜀ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

300 × 0.385 (285 – Tₑ) = 1000 × 4.184 (Tₑ – 26.5)

115.5(285 – Tₑ) = 4184(Tₑ – 26.5)

Clear the bracket

32917.5 – 115.5Tₑ = 4184Tₑ – 110876

Collect like terms

32917.5 + 110876 = 4184Tₑ + 115.5Tₑ

143793.5 = 4299.5Tₑ

Divide both side by 4299.5

Tₑ = 143793.5 / 4299.5

Tₑ = 33.4 °C

Therefore, the final temperature of water is 33.4 °C.

it's c (engineer's do their work with mechanical stuff not on any living being)