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Determine the latent heat of fusion of mercury using the
Chapter 14, Problem 31(choose chapter or problem)
Determine the latent heat of fusion of mercury using the following calorimeter data: 1.00 kg of solid Hg at its melting point of -39.0° C is placed in a 0.620-kg aluminum calorimeter with 0.400 kg of water at 12.80° C; the resulting equilibrium temperature is 5.06° C.
Questions & Answers
QUESTION:
Determine the latent heat of fusion of mercury using the following calorimeter data: 1.00 kg of solid Hg at its melting point of -39.0° C is placed in a 0.620-kg aluminum calorimeter with 0.400 kg of water at 12.80° C; the resulting equilibrium temperature is 5.06° C.
ANSWER:Step 1 of 2
Given data:
The mass of the solid Hg is given as \(m_1\) = 1 kg.
The mass of the aluminum container is given as \(m_2\) = 0.620 kg.
The mass of the water is given as \(m_3\) = 0.4 kg.
The temperature of Hg is given as \(t_{1}=-39^{\circ} \mathrm{C}\).
The temperature of aluminum and water is given as \(\mathrm{t}_{2}=12.8^{\circ} \mathrm{C}\).
The resultant temperature is given as \(t=5.06^{\circ} \mathrm{C}\).
Specific heat of Hg is given as \(\mathrm{C}_{1}=138 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\).
The specific heat of aluminum is given as \(\mathrm{c}_{2}=900 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\) .
The specific heat of water is given as \(\mathrm{c}_{3}=4186 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\).