Question

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?

Answer 1

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

Data obtained from the question

• Mass of cold lead (M) = 54.3 g
• Temperature of lead (T) = 384.4 K
• Temperature of water (Tᵥᵥ) = 291.2 K
• Equilibrium temperature (Tₑ) = 297.6 K
• Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gK
• Specific heat capacity of lead (C) = 0.128 J/gK
• Mass 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

Learn more about heat transfer:

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