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.
Answer:
Please elaborate. Maybe if you had actual answer choices
Explanation:
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