Answer:
82 °C
Explanation:
Let the specific heat capacity of the coffee be that of water which is 4.2 J/g °C.
Now, at the final temperature, heat gained by Aluminum spoon ,Q equals heat lost by coffee, Q'.
Q = -Q'
Q = m₁c₁(T₂ - T₁) where m₁ = mass of aluminum spoon = 45 g, c₁ = specific heat of aluminum = 0.88 J/g °C, T₁ = initial temperature of aluminum spoon = 24 °C and T₂ = final temperature of aluminum spoon.
Q' = m₂c₂(T₂ - T₃) where m₂ = mass of coffee = 180 g, c₂ = specific heat of coffee = 4.2 J/g °C, T₃ = initial temperature of coffee = 85 °C and T₂ = final temperature of coffee.
So, Q = -Q'
m₁c₁(T₂ - T₁) = -m₂c₂(T₂ - T₃)
Making T₂ subject of the formula, we have
m₁c₁T₂ - m₁c₁T₁ = -m₂c₂T₂ + m₂c₂T₃
m₁c₁T₂ + m₂c₂T₂ = m₂c₂T₃ + m₁c₁T₁
(m₁c₁ + m₂c₂)T₂ = m₂c₂T₃ + m₁c₁T₁
T₂ = (m₂c₂T₃ + m₁c₁T₁)/(m₁c₁ + m₂c₂)
substituting the values of the variables into the equation, we have
T₂ = (180 g × 4.2 J/g °C × 85 °C + 45 g × 0.88 J/g °C × 24 °C )/(45 g × 0.88 J/g °C + 180 g × 4.2 J/g °C)
T₂ = (64260 J + 950.4 J)/(39.6 J/°C + 756 J/°C)
T₂ = 65210.4 J/795.6 J/°C
T₂ = 81.96 °C
T₂ ≅ 82 °C
