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?

Question

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?

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Huyền Thanh 4 years 2021-07-28T21:55:22+00:00 1 Answers 19 views 0

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  1. thachthao
    0
    2021-07-28T21:56:29+00:00
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    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.

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