A sample of lead with a mass of 54.3 g is heated to a temperature of 384.4 K and placed in a container of water at 291.2 K. The final temperature of the lead and water is 297.6 K. Assuming no heat loss, what mass of water was in the container?

Answers

Answer 1

The mass of the water in the container given the data from the question is 22.5 g

Data obtained from the questionMass of cold lead (M) = 54.3 gTemperature of lead (T) = 384.4 KTemperature of water (Tᵥᵥ) = 291.2 KEquilibrium temperature (Tₑ) = 297.6 KSpecific heat capacity of the water (Cᵥᵥ) = 4.184 J/gKSpecific heat capacity of lead (C) = 0.128 J/gKMass of water (Mᵥᵥ) = ?

How to determine the mass of water

Heat loss = Heat gain

MC(T – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

54.3 × 0.128 (384.4 – 297.6) = Mᵥᵥ × 4.184(297.6 – 291.2)

6.9504 × 86.8 = Mᵥᵥ × 4.184 × 6.4

Divide both side by 4.184 × 6.4

Mᵥᵥ = (6.9504 × 86.8) / (4.184 × 6.4)

Mᵥᵥ = 22.5 g

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[tex]_{15}^{30}P\ --- > \ _{14}^{30} Si \ + \ ^{0}_{1} n[/tex]

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Answers

Answer:

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Explanation:

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Answers

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Answers

Answer:

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Explanation:

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Answers

Answer:

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Answer:

The answer would be "Yellow"

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Answer:

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Answers

Answer:

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