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The heat of fusion of mercury is \(2.72 \mathrm{cal} / \mathrm{g} .\) Calculate the quantity of energy transferred when 4.37 mol Hg freezes at a temperature of \(-39^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The energy transferred is 2384.29 cal.

Step by step solution

01

Convert Moles to Grams

First, we need to know the molar mass of mercury (Hg) to convert moles into grams. The molar mass of Hg is 200.59 g/mol. Multiply the number of moles by the molar mass:\[\text{Mass of } \text{Hg } = 4.37 \text{ mol} \times 200.59 \frac{\text{g}}{\text{mol}} = 876.58 \text{ g}\]
02

Calculate Energy Transferred

Use the heat of fusion to find the total energy transferred. The heat of fusion for mercury is given as 2.72 cal/g. Multiply the mass from Step 1 by this value to get the energy:\[\text{Energy transferred} = 876.58 \text{ g} \times 2.72 \frac{\text{cal}}{\text{g}} = 2384.29 \text{ cal}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Transfer
Understanding energy transfer is key when dealing with phase changes like freezing or melting. During such phase changes, energy is transferred from one phase to another. This energy does not change the temperature but instead facilitates the transformation of the state of matter. For example, when mercury freezes, it releases energy into its surroundings. This release of energy corresponds to the heat of fusion—the amount of energy needed to change mercury from a liquid to a solid at its freezing point without changing its temperature.

In our exercise, knowing that the heat of fusion for mercury is 2.72 cal/g, the focus is on calculating how much energy is released when all of the mercury turns from liquid to solid. Energy transfer calculations are an integral part of understanding thermal processes and thermodynamics. This concept is applied by multiplying the mass of mercury by its heat of fusion to find the total energy involved in the phase change.
Molar Mass Conversion
Converting moles to grams is a straightforward yet vital step in many calculations involving chemical reactions or phase changes. The molar mass is a conversion factor that bridges moles, which are ideal for stoichiometric calculations, with grams, which are practical for laboratory measurements and calculations.

For mercury (Hg), the molar mass is provided as 200.59 g/mol. This value means that one mole of mercury, which contains Avogadro's number of atoms, has a mass of 200.59 grams. In the step-by-step solution, we multiplied the number of moles (4.37 mol for mercury) by the molar mass to find the total mass in grams:
  • Calculate mass: 4.37 mol × 200.59 g/mol
  • Resulting in 876.58 grams of mercury
This conversion is crucial since the heat of fusion is given in calories per gram, so we need to know the substance's mass in grams to continue with our calculations.
Phase Change Calculations
Phase change calculations involve using specific heat values that apply directly to the amount of matter experiencing a state transformation. In our scenario, we are addressing the freezing of mercury, a process where the physical state changes, thus requiring specific energy calculations.

The simple formula used for phase change is:
  • Energy = Mass × Heat of Fusion
For the given exercise, the calculations are centered around understanding that when mercury freezes, it requires a measurable amount of energy per gram. The heat of fusion is precisely this measure: the amount of energy per gram needed to transition from liquid to solid.

With the mass of mercury calculated to be 876.58 grams and the heat of fusion known (2.72 cal/g), we calculate the total energy released by multiplying these two values:
  • Energy = 876.58 grams × 2.72 cal/g = 2384.29 calories
Such calculations not only help in understanding freezing or melting processes but are also fundamental in engineering and scientific applications where precise energy budgeting is essential.

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Most popular questions from this chapter

Determine the amount of reaction (in moles) that takes place for each process $$ 2 \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{~g}) $$ (a) \(2 \mathrm{~mol} \mathrm{O}_{2}\) reacts (b) \(0.115 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{3}\) forms (c) \(4.73 \mathrm{~g}\) NO reacts

A \(25.0-\mathrm{mL}\) sample of benzene at \(19.9{ }^{\circ} \mathrm{C}\) was cooled to its melting point, \(5.5^{\circ} \mathrm{C}\), and then frozen. Calculate how much heat transfer to the surroundings occurred in this process. (The density of benzene is \(0.880 \mathrm{~g} / \mathrm{mL} ;\) its specific heat capacity is \(1.74 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}\), and its enthalpy of fusion is \(127 \mathrm{~J} / \mathrm{g} .)\)

The specific heat capacity of copper is \(0.385 \mathrm{~J} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\), whereas it is \(0.128 \mathrm{~J} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\) for gold. Assume you place \(100 . \mathrm{g}\) of each metal, originally at \(25^{\circ} \mathrm{C},\) in a boiling water bath at \(100^{\circ} \mathrm{C}\). If energy is transferred to each metal at the same rate, determine which piece of metal will reach \(100^{\circ} \mathrm{C}\) first.

If you want to convert \(56.0 \mathrm{~g}\) ice (at \(\left.0^{\circ} \mathrm{C}\right)\) to water at \(75.0^{\circ} \mathrm{C},\) calculate how many grams of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\), you would have to burn to supply the energy to melt the ice and then warm it to the final temperature (at 1 bar).

Explain in your own words why it is useful in thermodynamics to distinguish a system from its surroundings.

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