Answer:
68.6 °C
Explanation:
From conservation of energy, the heat lost by acetone, Q = heat gained by aluminum, Q'
Q = Q'
Q = mL where Q = latent heat of vaporization of acetone, m = mass of acetone = 3.33 g and L = specific latent heat of vaporization of acetone = 518 J/g
Q' = m'c(θ₂ - θ₁) where m' = mass of aluminum = 44.0 g, c = specific heat capacity of aluminum = 0.9 J/g°C, θ₁ = initial temperature of aluminum = 25°C and θ₂ = final temperature of aluminum = unknown
So, mL = m'c(θ₂ - θ₁)
θ₂ - θ₁ = mL/m'c
θ₂ = mL/m'c + θ₁
substituting the values of the variables into the equation, we have
θ₂ = 3.33 g × 518 J/g/(44.0 g × 0.9 J/g°C) + 25 °C
θ₂ = 1724.94 J/(39.6 J/°C) + 25 °C
θ₂ = 43.56 °C + 25 °C
θ₂ = 68.56 °C
θ₂ ≅ 68.6 °C
So, the final temperature (in °C) of the metal block is 68.6 °C.
The final temperature of the metal block is 74.97°C
The specific heat of a substance is the required quantity of heat needed to raise the temperature of 1 gram of the substance by 1° C.
From the parameters given:
The number of moles of acetone is:
[tex]\mathbf{= 3.33 g \times \dfrac{mol}{58.08 \ mol}}[/tex]
= 0.0573 mol
At standard conditions, the heat of vaporization of acetone is:
[tex]\mathbf{\Delta H = 32.0 \ kJ/mol \times 0.0578 \ mol } \\ \\ \mathbf{\Delta H = 1.8496 \ kJ } \\ \\ \mathbf{ \Delta H = 1.85 \times 10^3 \ J}[/tex]
Given that:
The specific heat of the aluminum is = 0.903 J/g° C
The heat energy can be computed as:
q = msΔT
q = 41 g × 0.903 J/g° C × (x - 25°C)
Using the calorimetry principle, heat energy lost by metal = heat energy gained by acetone.
i.e.
[tex]\mathbf{q_{(acetone)} gain = q_{(metal)} lost }[/tex]
[tex]\mathbf{-1.85 \times 10^3 \ J = - 41 g \times 0.903 \ J/g^0 C \times ( x - 25^0 c) }[/tex]
[tex]\mathbf{1.85 \times 10^3 \ J = 41 g \times 0.903 \ J/g^0 C \times ( x - 25^0 c) }[/tex]
[tex]\mathbf{(x - 25 ^0 C) = \dfrac{1.85 \times 10^3 \ J }{ 41 g \times 0.903 \ J/g^0 C}}[/tex]
[tex]\mathbf{(x - 25 ^0 C) = 49.97^0 C}}[/tex]
[tex]\mathbf{x = 49.97^0 C+25 ^0 C}}[/tex]
x = 74.97 °C
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