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

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