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>
- 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) = ?
<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
Learn more about heat transfer:
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