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.

Answer 1

m₁ = 0.37 kg

Step-by-step 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