Answer:
<em>The rate at which energy is removed = 59.085 J</em>
Explanation:
The rate at which energy is removed = c₁m₁ΔΘ/t + c₂m₂ΔΘ/t................. Equation 1
Where c₁ = specific heat capacity of the aluminum, m₁ = mass of the aluminum, ΔΘ/t = rate of temperature decrease, c₂ = specific heat capacity of water, m₂ = mass of water.
<em>Given: m₁ = 214 g = (214/1000) kg = 0.214 kg, m₂ = 892 g = (892/1000) kg </em>
<em>m₂ = 0.892 kg, ΔΘ/t = 0.9 °C/min. = (0.9/60) °C/seconds. = 0.015 °C/seconds.</em>
<em>Constants: c₁ = 900 J/kg.°C, c₂ = 4200 J/kg.°C</em>
<em>Substituting these values into Equation 1,</em>
<em>The rate at which Energy is removed = (0.214×900×0.015) + (0.892×4200×0.015)</em>
<em>The rate at which energy is removed = 2.889 + 56.196</em>
<em>The rate at which energy is removed = 59.085 J</em>