Find the temperature of a mixture obtained by mixing 32 grams of water at a temperature of 22 degrees Celsius with 88 grams of water at a te

Question

Find the temperature of a mixture obtained by mixing 32 grams of water at a temperature of 22 degrees Celsius with 88 grams of water at a temperature of 75 degrees Celsius.

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Ngọc Hoa 3 years 2021-07-28T12:28:59+00:00 1 Answers 5 views 0

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  1. yenoanh
    0
    2021-07-28T12:30:48+00:00
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    Answer:

    60.87 °C

    Explanation:

    Applying,

    Heat lost = heat gain

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

    Equation 1 can futher be simplified to

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

    Where m’ = mass of the hot water, m = mass of the cold water, t₁ = initial temperature of the cold water, t₂ = initial temperature of the hot water, t₃ = temperature of the mixture.

    From the question,

    Given: m’ = 88 g, m = 32 g, t₁ = 22°C, t₂ = 75°C

    Substitute these values into equation 2

    88(75-t₃) = 32(t₃-22)

    6600-88t₃ = 32t₃-704

    32t₃+88t₃ = 6600+704

    120t₃ = 7304

    t₃ = 60.87 °C

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