/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Ethyl alcohol has a boiling poin... [FREE SOLUTION] | 91Ó°ÊÓ

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Ethyl alcohol has a boiling point of \(78.0^{\circ} \mathrm{C},\) a freezing point of \(-114^{\circ} \mathrm{C},\) a heat of vaporization of \(879 \mathrm{~kJ} / \mathrm{kg},\) a heat of fusion of \(109 \mathrm{~kJ} / \mathrm{kg},\) and a specific heat of \(2.43 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} .\) How much energy must be removed from \(0.510 \mathrm{~kg}\) of ethyl alcohol that is initially a gas at \(78.0^{\circ} \mathrm{C}\) so that it becomes a solid at \(-114^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
742.099 kJ of energy must be removed.

Step by step solution

01

Condense Ethyl Alcohol from Gas to Liquid

Calculate the energy required to change ethyl alcohol from gas to liquid at its boiling point. The formula for heat removal during condensation is \[Q = m \times L_v\]where \(Q\) is the heat removed,\(m = 0.510 \ \text{kg}\) is the mass of the alcohol, and\(L_v = 879 \ \text{kJ/kg}\) is the heat of vaporization.Substitute the values:\[Q = 0.510 \ \text{kg} \times 879 \ \text{kJ/kg} = 448.29 \ \text{kJ}\]So, 448.29 kJ of heat must be removed to condense the ethyl alcohol.
02

Cool Liquid Ethyl Alcohol from Boiling Point to Freezing Point

Calculate the energy required to cool the liquid from its boiling point to its freezing point. The formula for the heat removed is \[Q = m \times c \times \Delta T\]where \(Q\) is the heat removed,\(m = 0.510 \ \text{kg}\) is the mass,\(c = 2.43 \ \text{kJ/kg \cdot K}\) is the specific heat capacity, and\(\Delta T = 78.0^{\circ}C - (-114^{\circ}C) = 192^{\circ}C\) (temperature change).Substitute the values:\[Q = 0.510 \ \text{kg} \times 2.43 \ \text{kJ/kg \cdot K} \times 192 \ \text{K} = 238.219 \ \text{kJ}\]So, 238.219 kJ of heat must be removed to cool the ethyl alcohol.
03

Freeze Ethyl Alcohol from Liquid to Solid

Calculate the energy required to freeze the liquid ethyl alcohol at its freezing point.The formula for heat removed during freezing is \[Q = m \times L_f\]where \(Q\) is the heat removed,\(m = 0.510 \ \text{kg}\) is the mass, and\(L_f = 109 \ \text{kJ/kg}\) is the heat of fusion.Substitute the values:\[Q = 0.510 \ \text{kg} \times 109 \ \text{kJ/kg} = 55.59 \ \text{kJ}\]So, 55.59 kJ of heat must be removed to freeze the ethyl alcohol.
04

Calculate Total Energy Removed

Add the energy removed in each step to find the total energy removed.Total energy removed is calculated as follows:\[Q_{\text{total}} = Q_{\text{condensation}} + Q_{\text{cooling}} + Q_{\text{freezing}}\]Substituting the values from previous steps:\[Q_{\text{total}} = 448.29 \ \text{kJ} + 238.219 \ \text{kJ} + 55.59 \ \text{kJ} = 742.099 \ \text{kJ}\]Hence, a total of 742.099 kJ of energy must be removed.

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Key Concepts

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

Phase Change
Phase change refers to the transformation of a substance from one state of matter to another. Ethyl alcohol can exist as a solid, liquid, or gas and each transition between these phases involves energy changes.

For example, when ethyl alcohol changes from a gas to a liquid, it undergoes condensation. This process occurs at its boiling point of around \(78.0^{\circ}C\). Energy, in the form of heat, is removed from the substance, allowing the molecules to come closer and form the liquid phase. The key term here is the "heat of vaporization" (\(L_v\)), which is the amount of energy needed to convert 1 kg of a substance from liquid to gas or vice versa. In our case, ethyl alcohol has a heat of vaporization of \(879 \, \text{kJ/kg}\).

Similarly, when the substance changes from liquid to solid, it goes through freezing, which happens at ethyl alcohol's freezing point, \(-114^{\circ}C\). The "heat of fusion" (\(L_f\)) is the energy required to change 1 kg of a substance from solid to liquid or vice versa, and for ethyl alcohol, it's \(109 \, \text{kJ/kg}\). Knowing these properties helps us calculate how much energy needs to be removed during phase changes.
Heat Transfer
Heat transfer is a fundamental principle responsible for the phase changes of substances. It's the process by which heat energy is transported from one area to another, often resulting in a phase change. This movement of heat happens due to the temperature difference between substances or within a substance.

In the case of ethyl alcohol, several heat transfer processes occur when going from a gas at \(78.0^{\circ}C\) to a solid at \(-114^{\circ}C\). By removing heat, we cause the liquid molecules to lose energy and slow down enough to transition into the solid phase.

Heat transfer calculations involve specific formulas like the heat equation for condensation:
  • \( Q = m \times L_v \)
Likewise, we calculate the heat transfer needed to cool the substance as it stays in one phase using:
  • \( Q = m \times c \times \Delta T \)
Accurate heat transfer calculations are critical for understanding how much energy must be added or removed in various scenarios.
Specific Heat Capacity
Specific heat capacity is a property of a substance that indicates how much heat energy is required to change the temperature of 1 kg of the substance by 1 Kelvin (or 1 degree Celsius). For ethyl alcohol, the specific heat capacity is \(2.43 \, \text{kJ/kg} \cdot \text{K}\).

This property is essential when you want to change the temperature of a substance without changing its phase, as it tells you how energy efficient the material is in storing heat.

In practical problems, specific heat capacity is used in calculations like:
  • \( Q = m \times c \times \Delta T \)
where \(Q\) is the heat added or removed, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change. Knowing the specific heat capacity allows us to understand how much energy is necessary to bring ethyl alcohol down from \(78.0^{\circ}C\) to its freezing point of \(-114^{\circ}C\) without it changing phase.

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Most popular questions from this chapter

A recruit can join the semi-secret "300 F" club at the Amundsen-Scott South Pole Station only when the outside temperature is below \(-70^{\circ} \mathrm{C}\). On such a day, the recruit first basks in a hot sauna and then runs outside wearing only shoes. (This is, of course, extremely dangerous, but the rite is effectively a protest against the constant danger of the cold. Assume that upon stepping out of the sauna, the recruit's skin temperature is \(102^{\circ} \mathrm{F}\) and the walls, ceiling, and floor of the sauna room have a temperature of \(30^{\circ} \mathrm{C}\). Estimate the recruit's surface area, and take the skin emissivity to be \(0.80 .\) (a) What is the approximate net rate \(P_{\text {net }}\) at which the recruit loses energy via thermal radiation exchanges with the room? Next, assume that when outdoors, half the recruit's surface area exchanges thermal radiation with the sky at a temperature of \(-25^{\circ} \mathrm{C}\) and the other half exchanges thermal radiation with the snow and ground at a temperature of \(-80^{\circ} \mathrm{C}\). What is the approximate net rate at which the recruit loses energy via thermal radiation exchanges with (b) the sky and (c) the snow and ground?

On finding your stove out of order, you decide to boil the water for a cup of tea by shaking it in a thermos flask. Suppose that you use tap water at \(19^{\circ} \mathrm{C},\) the water falls \(32 \mathrm{~cm}\) each shake, and you make 27 shakes each minute. Neglecting any loss of thermal energy by the flask, how long (in minutes) must you shake the flask until the water reaches \(100^{\circ} \mathrm{C} ?\)

In a certain solar house, energy from the Sun is stored in barrels filled with water. In a particular winter stretch of five cloudy days, \(1.00 \times 10^{6}\) kcal is needed to maintain the inside of the house at \(22.0^{\circ} \mathrm{C}\). Assuming that the water in the barrels is at \(50.0^{\circ} \mathrm{C}\) and that the water has a density of \(1.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3},\) what volume of water is required?

An object of mass \(6.00 \mathrm{~kg}\) falls through a height of \(50.0 \mathrm{~m}\) and, by means of a mechanical linkage, rotates a paddle wheel that stirs \(0.600 \mathrm{~kg}\) of water. Assume that the initial gravitational potential energy of the object is fully transferred to thermal energy of the water, which is initially at \(15.0^{\circ} \mathrm{C}\). What is the temperature rise of the water?

In the extrusion of cold chocolate from a tube, work is done on the chocolate by the pressure applied by a ram forcing the chocolate through the tube. The work per unit mass of extruded chocolate is equal to \(p / \rho,\) where \(p\) is the difference between the applied pressure and the pressure where the chocolate emerges from the tube, and \(\rho\) is the density of the chocolate. Rather than increasing the temperature of the chocolate, this work melts cocoa fats in the chocolate. These fats have a heat of fusion of \(150 \mathrm{~kJ} / \mathrm{kg} .\) Assume that all of the work goes into that melting and that these fats make up \(30 \%\) of the chocolate's mass. What percentage of the fats melt during the extrusion if \(p=5.5 \mathrm{MPa}\) and \(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3} ?\)

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