Ask Question
3 October, 08:02

A hot lump of 27.7 g of aluminum at an initial temperature of 58.8 °C is placed in 50.0 mL H 2 O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water, given that the specific heat of aluminum is 0.903 J / (g⋅°C) ? Assume no heat is lost to surroundings.

+1
Answers (1)
  1. 3 October, 10:25
    0
    The final temperature of the aluminium and water = 28.61 °C

    Explanation:

    Heat lost by the aluminum = heat gained by the water.

    c₁m₁ (t₁ - t₃) = c₂m₂ (t₃ - t₂) ... Equation 1

    making t₃ the subject of formula in equation 1

    t₃ = (c₁m₁t₁ + c₂m₂t₂) / (c₂m₂+c₁m₁) ... Equation 2

    Where c₁ = specific heat capacity of aluminium, m₁ = mass of aluminium, c₂ = specific heat capacity of water, m₂ = mass of water, t₁ = initial temperature of aluminum, t₂ = initial temperature of water, t₃ = temperature of the mixture.

    Given: c₁ = 0.903 J/g.°C, m₁ = 27.7 g, t₁ = 58.8 °C, t₂ = 25 °C

    mass = density * Volume

    Density of water at 25 °C = 0.997 g/mL

    Volume of water = 50.0 mL

    ∴ m₂ = 50 * 0.997 = 49.85 g

    m₂ = 49.85 g.

    Constant: c₂ = 4.2 J/g.°C

    Substituting these values into equation 2

    t₃ = [ (0.903*27.7*58.80) + (4.2*49.85*25) ]/[ (0.903*27.7) + (4.2*49.85) ]

    t₃ = (1470.77 + 5234.25) / (25.01+209.37)

    t₃ = 6705.02/234.38

    t₃ = 28.61 °C

    Therefore the final temperature of the aluminium and water = 28.61 °C
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A hot lump of 27.7 g of aluminum at an initial temperature of 58.8 °C is placed in 50.0 mL H 2 O initially at 25.0 °C and allowed to reach ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers