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
temperature of the mixture will be 75°C? For water, c = 4.19 x 103 J/(kg • K).
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Answer 1

Answer:

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

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