Suppose that 0.52 g of water at 25 ∘c condenses on the surface of a 51-g block of aluminum that is initially at 25 ∘c. if the heat released during condensation goes only toward heating the metal, what is the final temperature (in ∘c) of the metal block

According to the Law of Conservation of Energy, energy can neither be created nor destroyed. It is just transferred from one system to each other in different forms. In this case, the heat released by the water through condensation, is used to heat the metal block. 

Heat of water = Heat of aluminum

To determine the heat released during a phase change, we focus on the latent heat. The latent heat of condensation is equal to -2,260 J/g. Hence,

Heat of water = (0.52 g)(-2,260 J/g) = -1,175.2 J

The heat of aluminum is sensible heat. It is equal to the mass times specific heat of the aluminum times the temperature change: H = mCpΔT. The specific heat of aluminum is 0.921 J/g-°C.

Heat of aluminum = (51 g)(0.921 J/g-°C)(T - 25°C) = -1,175.2 J

Solving for T,
T = 50°C

Therefore, the final temperature is 50°C.

The concentration of Mg²⁺ cation at end point: M₂ = 0.0048 M

Explanation:

The titration reaction involved: MgCl₂(aq) + 2 NaOH(aq) → Mg(OH)₂(s) + 2 NaCl(aq)

and the solubility equilibrium for the product: Mg(OH)₂(s) ⇌ Mg²⁺ (aq) + 2 OH⁻(aq)

Given: Concentration of MgCl₂: M₁ = 0.005 M, volume of MgCl₂: V₁ = 100 mL, Total volume of solution at end point = 104.12 mL

Concentration of Mg²⁺ cation at end point: M₂ = ?M

Now, to calculate the concentration of Mg²⁺ cation at end point, we use the equation:    M₁ × V₁ = M₂ × V₂

⇒ M₂ = M₁ × V₁ ÷ V₂

⇒ M₂ = (0.005 M × 100 mL) ÷ 104.12 mL

⇒ M₂ = 0.0048 M (2 significant digits)

Therefore, the concentration of Mg²⁺ cation at end point: M₂ = 0.0048 M