When 155 g of water at 20 ᵒC is mixed with 75 g of water at 80 ᵒC, assuming no heat is lost to the surrounding what is the final

Question

When 155 g of water at 20 ᵒC is mixed with 75 g of water at 80 ᵒC, assuming no heat is

lost to the surrounding what is the final temperature of the water. (Specific heat capacity

of water is 4.18 J/°C·g)​

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Bơ 5 years 2021-08-04T20:36:52+00:00 1 Answers 255 views 1

Answers ( )

  1. Nick
    1
    2021-08-04T20:38:20+00:00
    Reply

    Answer:

    39.57 °C

    Explanation:

    Applying,

    Heat gain = Heat lost.

    cm'(t₃-t₁) = cm(t₂-t₃)…………….. Equation 1

    Where c = specific heat capacity of water, m = mass of the hotter water, m’ = mass of the colder water, t₁ = initial temperature of the colder water, t₂ = initial temperature of the hotter water, t₃ = Final Temperature.

    Equation 1 above can futher be simplified to

    m'(t₃-t₁) = m(t₂-t₃)…………….. Equation 2

    From the question,

    Given: m’ = 155 g, m’ = 75 g, t₁ = 20 °C, t₂ = 80 °C

    Substitute these values into equation 2

    155(t₃-20) = 75(80-t₃)

    Solve for t₃

    155t₃- 3100 = 6000-75t₃

    155t₃+75t₃ = 6000+3100

    230t₃ = 9100

    t₃ = 9100/230

    t₃ = 39.57 °C

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