Question

The value of specific heat for copper is 390 J/kg⋅C∘, for aluminun is 900 J/kg⋅C∘, and for water is 4186 J/kg⋅C∘.

Part A

What will be the equilibrium temperature when a 275 g block of copper at 245 ∘C is placed in a 135 g aluminum calorimeter cup containing 855 g of water at 15.0 ∘C?

Express your answer using three significant figures.'

T=  ∘C

Answer 1

Let’s convert the masses to kilograms.
Cu = 0.275, Al = 0.135, H2O = 0.855

The initial temperature of the aluminum and water is 15˚. When the copper is placed in the cup, the temperature of the copper will decrease as the temperature of the aluminum and water increases. Use the following equation to determine the heat released by the copper.

Q = m * Specific heat * (Ti – Tf)
Q = 0.275 * 390 * (245 – Tf) = 26276.25 – 107.25 * Tf

Use the following equation to determine the heat absorbed by the aluminum and water.
Q = m * Specific heat * (Tf – Ti)

For Al, Q = 0.135 * 900 * (Tf – 15) = 121.5 * Tf – 1822.5
For H2O, Q = 0.855 * 4186 * (Tf – 16) = 3579.03* Tf – 57264.48
Total heat absorbed = 121.5 * Tf – 1822.5 + 3579.03* Tf – 57264.48 = 3700.53 * Tf – 55441.98

Total heat absorbed = heat released

3700.53* Tf – 55441.98 = 26276.25 – 107.25* Tf
3807.78 * Tf = 81718.23
Tf = 81718.23÷ 3807.78 = 21.46˚