Question

A 45g Aluminum spoon (specific heat 0.88J/g degree Celcius) at 24 degrees Celcius placed in 180ml(180g) of coffee at 85 degrees Celcius and the temperature of the two become equal. What is the final temperature?

Answer 1

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