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Chapter 18: Problem 39

co 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
741.26 kJ must be removed.

Step by step solution

01

Calculate the energy to condense the gas

First, we need to remove the heat from the ethyl alcohol gas at its boiling point to condense it into a liquid. The formula for this is:\[ Q_1 = m \times L_v \]where \( m = 0.510 \text{ kg} \) is the mass of ethyl alcohol and \( L_v = 879 \text{ kJ/kg} \) is the heat of vaporization.Substituting the values, we get:\[ Q_1 = 0.510 \times 879 = 448.29 \text{ kJ} \]
02

Calculate the energy to cool the liquid to the freezing point

Next, we cool the liquid ethyl alcohol from its boiling point (\(78.0^{\circ} \text{C}\)) to its freezing point (\(-114^{\circ} \text{C}\)). The formula for calculating this energy is:\[ Q_2 = m \times c \times \Delta T \]where \( c = 2.43 \text{ kJ/kg} \cdot \text{K} \) is the specific heat capacity, and \( \Delta T = 78 - (-114) = 192 \text{ K} \) is the change in temperature.Substituting the values, we get:\[ Q_2 = 0.510 \times 2.43 \times 192 = 237.38 \text{ kJ} \]
03

Calculate the energy to freeze the liquid into a solid

Finally, we calculate the energy required to freeze the ethyl alcohol at its freezing point using the formula:\[ Q_3 = m \times L_f \]where \( L_f = 109 \text{ kJ/kg} \) is the heat of fusion.Substituting the values, we get:\[ Q_3 = 0.510 \times 109 = 55.59 \text{ kJ} \]
04

Sum all energy changes

To find the total energy that must be removed, we sum the energy changes from each step:\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 \]Substituting the values from previous steps, we have:\[ Q_{\text{total}} = 448.29 + 237.38 + 55.59 = 741.26 \text{ kJ} \]

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Key Concepts

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

Phase Transition
Phase transitions occur when a substance changes from one state of matter (solid, liquid, or gas) to another. This transition is caused by changes in temperature or pressure, which alter the energy within the substance. During a phase transition, the energy absorbed or released does not change the temperature, but instead is used to alter the state of the substance.
For example:
  • Melting is when a solid turns into a liquid.
  • Freezing is when a liquid turns into a solid.
  • Evaporation or boiling is when a liquid turns into a gas.
  • Condensation is when a gas turns into a liquid.
Understanding these processes is crucial in thermodynamics, as it helps explain how energy is transferred within different systems.
Heat of Vaporization
The heat of vaporization is the amount of energy required to turn a liquid into a gas at its boiling point. It is a vital property in thermodynamics, as it indicates how much energy is needed to break intermolecular bonds during the conversion from liquid to vapor.
For ethyl alcohol, the heat of vaporization is 879 kJ/kg. This value signifies the energy per kilogram that must be removed for ethyl alcohol to condense back into liquid when it's a gas.
  • This process requires energy to overcome the forces keeping the liquid particles together.
  • The energy measured as the heat of vaporization is large because changing a liquid to vapor requires breaking many intermolecular forces.
This principle is essential in applications like refrigeration and distillation.
Specific Heat Capacity
Specific heat capacity is the amount of heat energy required to raise the temperature of 1 kilogram of a substance by 1 Kelvin (K). It is a unique feature of different materials and influences how they retain and transfer heat.
Ethyl alcohol has a specific heat capacity of 2.43 kJ/kg路K. This indicates how well ethyl alcohol can store thermal energy when changing temperature without changing state.
  • A higher specific heat capacity means the substance can absorb a lot of heat before its temperature changes by 1 K.
  • In the problem, calculating how much energy ethyl alcohol needs to go from its boiling point to its freezing point uses this value.
This property is crucial for designing processes that require precise thermal management, such as cooling and heating systems.
Enthalpy Change
Enthalpy change refers to the total heat content change in a system during a process, like a phase transition. It represents the energy absorbed or released under constant pressure, reflecting the system's internal energy changes.
In our exercise, calculating the full enthalpy change involves understanding each phase transition step:
  • Condensing the ethyl alcohol gas into a liquid.
  • Cooling the liquid to its freezing point.
  • Freezing the liquid into a solid.
Each part has its own energy change, which combines to give the entire transformation鈥檚 enthalpy change. This calculation helps in predicting how systems respond to energy changes, making it fundamental in fields like chemistry and physics.

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Most popular questions from this chapter

The average rate at which en- Fig. 18-51 Problem 71. ergy is conducted outward through the ground surface in North America is \(54.0 \mathrm{~mW} / \mathrm{m}^{2}\), and the average thermal conductivity of the near-surface rocks is \(2.50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming a surface temperature of \(10.0^{\circ} \mathrm{C}\), find the temperature at a depth of \(35.0 \mathrm{~km}\) (near the base of the crust). Ignore the heat generated by the presence of radioactive elements.

\(\$$ An insulated Thermos contains \)130 \mathrm{~cm}^{3}\( of hot coffee at \)80.0^{\circ} \mathrm{C}\(. You put in a \)12.0 \mathrm{~g}$ ice cube at its melting point to cool the coffee. By how many degrees has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment.

(c) Evaporative cooling of beverages. A cold beverage can be kept cold even on a warm day if it is slipped into a porous ceramic container that has been soaked in water. Assume that energy lost to evaporation matches the net energy gained via the radiation exchange through the top and side surfaces. The container and beverage have temperature \(T=15^{\circ} \mathrm{C}\), the environment has temperature \(T_{\text {env }}=32^{\circ} \mathrm{C}\), and the container is a cylinder with radius \(r=2.2 \mathrm{~cm}\) and height \(10 \mathrm{~cm}\). Approximate the emissivity as \(\varepsilon=1\), and neglect other energy exchanges. At what rate \(d m / d t\) is the container losing water mass?

Suppose \(200 \mathrm{~J}\) of work is done on a system and \(70.0 \mathrm{cal}\) is extracted from the system as heat. In the sense of the first law of thermodynamics, what are the values (including algebraic signs) of \((\mathrm{a}) W,(\mathrm{~b}) Q\), and \((\mathrm{c}) \Delta E_{\mathrm{int}} ?\)

A \(2.50 \mathrm{~kg}\) lump of aluminum is heated to \(92.0^{\circ} \mathrm{C}\) and then dropped into \(8.00 \mathrm{~kg}\) of water at \(5.00^{\circ} \mathrm{C}\). Assuming that the lump-water system is thermally isolated, what is the system's equilibrium temperature?
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