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

Question

2 years ago

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final temperature of the lead and water is 297.6 K. Assuming no heat loss, what mass of water was in the container?

1 answer:

DerKrebs [107] · Answer 1 · 2023-06-14T22:58:03+03:00

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

<h3>Data obtained from the question</h3>

<h3>How to determine the mass of water </h3>

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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