Ask Question
22 August, 06:55

You have 750 g of water at 10°C in a large insulated beaker. How much boiling water at 100°C must you add to this beaker so that the final

temperature of the mixture will be 75°C? For water, c = 4.19 x 103 J / (kg • K).

naman ang kanilang a movementvou

+1
Answers (1)
  1. 22 August, 09:59
    0
    Correct answer: m₂ = 1,950 g

    Explanation:

    When mixing the liquid, in this case water, at different temperatures, the warmer liquid will cool and the cooler will warm up until their temperatures equalize.

    The amount of heat that the second fluid (water) releases is equal to the amount of heat that the first fluid (water) receives in this spontaneous process.

    Q₁ = Q₂

    Q₁ = m₁ c (t - t₁) and Q₂ = m₂ c (t₂ - t)

    where are they: m₁ = 750 g, t₁ = 10°C, t₂ = 100°C, t = 75°C, m₂ = ?

    m₁ c (t - t₁) = m₂ c (t₂ - t) / : c

    Since water is the liquid in question, it has the same specific heat capacity. Therefore, we will divide both sides of the equation by c and get:

    m₁ (t - t₁) = m₂ (t₂ - t) = > m₂ = m₁ (t - t₁) / (t₂ - t) = >

    m₂ = 750 · (75 - 10) / (100 - 75) = > m₂ = 750 · 65 / 25 = 1,950 g

    m₂ = 1,950 g

    God is with you!
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “You have 750 g of water at 10°C in a large insulated beaker. How much boiling water at 100°C must you add to this beaker so that the final ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers