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Ethyl chloride \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) boils at \(12^{\circ} \mathrm{C}\). When liquid \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) under pressure is sprayed on a room-temperature \(\left(25^{\circ} \mathrm{C}\right)\) surface in air, the surface is cooled considerably. (a) What does this observation tell us about the specific heat of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(g)\) as compared with that of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(l) ?(\mathbf{b})\) Assume that the heat lost by the surface is gained by ethyl chloride. What enthalpies must you consider if you were to calculate the final temperature of the surface?

Short Answer

Expert verified
The observation indicates that the specific heat of ethyl chloride in its gaseous state, \(C_{g}\), must be greater than the specific heat of ethyl chloride in its liquid state, \(C_{l}\). To calculate the final temperature of the surface, we must consider the: 1. Enthalpy of vaporization, 2. Enthalpy of heating, and 3. Enthalpy of cooling, by accounting for their respective effects on temperature.

Step by step solution

01

Analyze the Observation

The observation tells us that when liquid ethyl chloride is sprayed on a room-temperature surface, the surface cools down considerably. This indicates that the gaseous ethyl chloride has a lower temperature than the liquid ethyl chloride, which means that the heat is being transferred from the surface to the ethyl chloride.
02

Compare Specific Heat of Ethyl Chloride in Liquid and Gaseous States

From the observation, we can infer that the specific heat of ethyl chloride in its gaseous state, \(C_{g}\), must be greater than the specific heat of ethyl chloride in its liquid state, \(C_{l}\). This is because a higher specific heat allows the gaseous ethyl chloride to absorb more heat per unit mass, resulting in a significant temperature decrease on the surface.
03

Identify the Enthalpies Involved in the Process

To calculate the final temperature of the surface, we must consider the following enthalpies: 1. Enthalpy of vaporization: The heat absorbed by liquid ethyl chloride to convert into a gaseous state. 2. Enthalpy of heating: The heat absorbed by gaseous ethyl chloride to increase its temperature. 3. Enthalpy of cooling: The heat lost by the surface to cool down. These enthalpies are related to each other by the principle of conservation of energy: the heat lost by the surface must be equal to the heat gained by the ethyl chloride. To calculate the final temperature of the surface, we need to account for all these enthalpies and their respective effects on temperature.

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Key Concepts

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

Specific Heat
Specific heat refers to the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. It is an intrinsic property that varies among different substances and states of matter. In our scenario with ethyl chloride, an important observation is made. When this substance is sprayed on a surface at room temperature, the surface experiences a cooling effect.
This suggests that the specific heat of ethyl chloride in its gaseous state,
  • Gaseous specific heat ( $C_g$ ) is higher than its liquid specific heat ( $C_l$ ). This means $C_g > C_l$.
  • A higher specific heat implies that the ethyl chloride gas can absorb more heat without a large temperature change compared to its liquid form.
The capacity of gaseous ethyl chloride to absorb more heat leads to a significant lowering of the surface's temperature when sprayed, making this property a telling indicator of its specific heat behavior.
Enthalpy of Vaporization
The enthalpy of vaporization plays a crucial role in the cooling process we observe. This is the amount of energy needed to convert a liquid into a gas at a constant temperature. When ethyl chloride transitions from liquid to gas while sprayed, it absorbs energy in the form of heat from its surroundings, which, in this case, is the surface. To understand what's happening:
  • The liquid ethyl chloride requires energy to overcome molecular attractions to become gas. This is captured by its enthalpy of vaporization,
    • Represented by \(\Delta H_{vap}\), where ethyl chloride requires considerable heat to evaporate compared to less volatile liquids.
  • As a result, the energy absorbed during this phase change triggers the cooling of the surface.
This phase change process is at the heart of many cooling systems, making the enthalpy of vaporization a key concept in understanding heat transfer mechanisms.
Heat Transfer
Heat transfer is the movement of thermal energy from one body to another. In the context of the ethyl chloride scenario, the heat transfer results in cooling. Let's examine how this process unfolds:
  • When heat is lost from the surface, it's primarily because the sprayed ethyl chloride absorbs it during vaporization.
  • The fundamental principle at play here is the conservation of energy, where the energy lost by the surface is gained by ethyl chloride.
  • The surface's heat transfer includes the energy responsible for the liquid to gas transition and subsequent heating of the vapor.
Understanding this heat transfer helps to recognize the temperature changes, highlighted by the notable heat exchange between the surface and the ethyl chloride.

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Most popular questions from this chapter

(a) Which is generally stronger, intermolecular interactions or intramolecular interactions? (b) Which of these kinds of interactions are broken when a liquid is converted to a gas?

(a) When you exercise vigorously, you sweat. How does this help your body cool? (b) A flask of water is connected to a vacuum pump. A few moments after the pump is turned on, the water begins to boil. After a few minutes, the water begins to freeze. Explain why these processes occur.

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At standard temperature and pressure, the molar volumes of \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) gases are 22.06 and \(22.40 \mathrm{~L}\), respectively. (a) Given the different molecular weights, dipole moments, and molecular shapes, why are their molar volumes nearly the same? (b) On cooling to \(160 \mathrm{~K}\), both substances form crystalline solids. Do you expect the molar volumes to decrease or increase on cooling the gases to \(160 \mathrm{~K} ?\) (c) The densities of crystalline \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) at \(160 \mathrm{~K}\) are 2.02 and \(0.84 \mathrm{~g} / \mathrm{cm}^{3}\), respectively. Calculate their molar volumes. (d) Are the molar volumes in the solid state as similar as they are in the gaseous state? Explain. (e) Would you expect the molar volumes in the liquid state to be closer to those in the solid or gaseous state?

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