You place a piece of aluminum at 250.0∘C ∘ C in 9.00 kg k g of liquid water at 20.0∘C ∘ C . None of the water boils, and the final temperatu

Question

You place a piece of aluminum at 250.0∘C ∘ C in 9.00 kg k g of liquid water at 20.0∘C ∘ C . None of the water boils, and the final temperature of the water and aluminum is 22.0∘C ∘ C . What is the mass of the piece of aluminum? Assume no heat is exchanged with the container that holds the water.

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Hải Đăng 6 months 2021-07-21T15:55:39+00:00 1 Answers 6 views 0

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    2021-07-21T15:57:07+00:00

    Answer:

    m₁ = 0.37 kg

    Explanation:

    According to Law of conservation of energy:

    Heat Lost by Aluminum = Heat Gained by Water

    m₁C₁ΔT₁ = m₂C₂ΔT₂

    where,

    m₁ = mass of piece of aluminum = ?

    C₁ = specific heat capacity of aluminum = 900 J/kg.°C

    ΔT₁ = Change in temperature of aluminum = 250°C – 22°C = 228°C

    m₂ = mass of water = 9 kg

    C₂ = specific heat capacity of water = 4200 J/kg.°C

    ΔT₁ = Change in temperature of aluminum = 22°C – 20°C = 2°C

    Therefore,

    m₁(900 J/kg.°C)(228 °C) = (9 kg)(4200 J/kg.°C)(2°C)

    m₁ = (75600 J)/(205200 J/kg)

    m₁ = 0.37 kg

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