Ahot 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.

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

it has to be b, about 97 percent.