### Solution Found!

# Determine the latent heat of fusion of mercury using the

**Chapter 14, Problem 31**

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**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.

### 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 \(\mathrm{c}_{3}=4186 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\).