Answer:
60 °C
Explanation:
From the question given above, the following data were obtained:
Mass of 1st sample (M₁) = 200 g
Temperature of 1st sample (T₁) = 80 °C
Mass of 2nd sample (M₂) = 100 g
Temperature of 2nd sample (T₂) = 20 °C
Equilibrium temperature (Tₑ) =?
NOTE: Since the sample are the same, the specific heat capacity is constant.
We can obtain the equilibrium temperature as follow:
Heat lost by 1st = heat gained by the 2nd
M₁C(T₁ – Tₑ) = M₂C(Tₑ – T₂)
Cancel out C
M₁(T₁ – Tₑ) = M₂(Tₑ – T₂)
200 (80 – Tₑ) = 100 (Tₑ – 20)
Clear bracket
16000 – 200Tₑ = 100Tₑ – 2000
Collect like terms
16000 + 2000 = 100Tₑ + 200Tₑ
18000 = 300Tₑ
Divide both side by 300
Tₑ = 18000 / 300
Tₑ = 60 °C
Therefore, the equilibrium temperature (i.e the final temperature) of the mixture is 60 °C.
