Answer:
The final temperature of the aluminium and water = 28.61 °C
Step-by-step 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