A (200 g) of water at (80 °C) is mixed with (100 g)of water at (20 °C). What is the final temperature of the water?

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.

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