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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
740.91 kJ of energy must be removed.

Step by step solution

01

Condensation of Ethyl Alcohol

To calculate the energy removed during condensation, we use the formula: \( Q_1 = m \times L_v \), where \( Q_1 \) is the energy removed, \( m = 0.510 \text{ kg} \) is the mass, and \( L_v = 879 \text{ kJ/kg} \) is the heat of vaporization. Substituting the given values, \( Q_1 = 0.510 \times 879 = 448.29 \text{ kJ} \).
02

Cooling Liquid Ethyl Alcohol

Once the ethyl alcohol has condensed into a liquid at \(78.0^{\circ} \text{C}\), the next step is to cool it to its freezing point, \(-114^{\circ} \text{C}\). However, there is an error here as the liquid needs to cool to its freezing point at \(-114^{\circ} \text{C}\). Given the specific heat \(c = 2.43 \text{ kJ/kg} \cdot \text{K}\), we use the formula \( Q_2 = m \times c \times \Delta T \), where \( \Delta T = 192^{\circ}\text{C} \) (from \(78^{\circ}\text{C} \) to \(-114^{\circ} \text{C} \)). Calculate: \( Q_2 = 0.510 \times 2.43 \times 192 = 237.0276 \text{ kJ} \).
03

Freezing the Liquid to Solid

To calculate the energy removed during freezing, use the formula: \( Q_3 = m \times L_f \), where \( Q_3 \) is the energy removed, \( L_f = 109 \text{ kJ/kg} \) is the heat of fusion. Substituting the given values, \( Q_3 = 0.510 \times 109 = 55.59 \text{ kJ} \).
04

Total Energy Removal

The total energy removed is the sum of the energies removed in each step: \( Q_{total} = Q_1 + Q_2 + Q_3 \). Thus, \( Q_{total} = 448.29 + 237.0276 + 55.59 = 740.9076 \text{ kJ} \).

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Key Concepts

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

Phase Change
When substances change from one state of matter to another, such as from liquid to gas or gas to solid, it's called a phase change. These processes occur at specific temperatures, known as melting points, freezing points, or boiling points. Notably, in our exercise, ethyl alcohol transitions from a gaseous state at its boiling point of \(78.0^{\circ} \mathrm{C}\) to a solid state at \(-114^{\circ} \mathrm{C}\).

Phase changes require or release energy, described as latent heat. For instance:
  • **Heat of Vaporization** - Energy needed to convert a liquid to a gas. In ethyl alcohol, this is \(879 \mathrm{~kJ} / \mathrm{kg}\), which is significant during the condensation phase.
  • **Heat of Fusion** - Energy required to change a solid to a liquid, or vice versa. For our exercise, ethyl alcohol's heat of fusion is \(109 \mathrm{~kJ} / \mathrm{kg}\), essential when freezing the liquid alcohol into solid form.
Understanding these phase changes helps calculate the total energy needed to alter the state of a given mass, like the 0.510 kg of ethyl alcohol in our problem.
Heat Transfer
Heat transfer refers to the movement of thermal energy from one place to another. In the context of our exercise, heat is transferred out of the ethyl alcohol to change its state and temperature. The process includes several stages:

  • During **condensation**, energy is removed to transition from gas to liquid.
  • While **cooling** the liquid ethyl alcohol down to its freezing point, more energy must be taken out.
  • Finally, during **freezing**, additional energy is released as the liquid turns into a solid.
Each stage of this transformation relies on the principles of heat transfer, ensuring energy is properly calculated for the phase changes. These calculations use constants like the heat of vaporization and heat of fusion, which determine how much energy is needed or released as the alcohol changes state. Proper heat transfer calculations are crucial in thermodynamics, helping to design processes and systems like refrigerators, engines, and even climate control systems.
Specific Heat
Specific heat is an important concept in thermodynamics. It represents the amount of heat energy required to raise the temperature of one kilogram of a substance by one Kelvin. For ethyl alcohol in our exercise, the specific heat is \(2.43 \mathrm{~kJ/kg} \cdot \mathrm{K}\). This value helps in calculating how much energy must be removed to cool the liquid alcohol from its boiling point to its freezing point.

To determine the energy requirement for cooling, we use the formula:
\[ Q = m \times c \times \Delta T \]
Where:
  • **Q** is the energy transferred,
  • **m** is the mass of the substance (0.510 kg in this case),
  • **c** is the specific heat,
  • **\(\Delta T \)** is the change in temperature (from \(78^{\circ}\text{C} \) to \(-114^{\circ}\text{C}\), which is a total of \(192^{\circ}\text{C}\)).
This formula explains how specific heat is used to quantify energy transfer needed for temperature changes, excluding any phase change. It ensures accurate calculations in cooling and heating processes across various scientific and engineering applications.

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Most popular questions from this chapter

Suppose the temperature of a gas is \(373.15 \mathrm{~K}\) when it is at the boiling point of water. What then is the limiting value of the ratio of the pressure of the gas at that boiling point to its pressure at the triple point of water? (Assume the volume of the gas is the same at both temperatures.)

Leidenfrost effect. \(\mathrm{A}\) water drop will last about \(1 \mathrm{~s}\) on a hot skillet with a temperature between \(100^{\circ} \mathrm{C}\) and about \(200^{\circ} \mathrm{C}\). However, if the skillet is much hotter, the drop can last several minutes, an effect named after an early investigator. The longer lifetime is due to the support of a thin layer of air and water vapor that separates the drop from the metal (by distance \(L\) in Fig. \(18-48)\). Let \(L=\) \(0.100 \mathrm{~mm}\), and assume that the drop is flat with height \(h=1.50 \mathrm{~mm}\) and bottom face area \(A=4.00 \times 10^{-6} \mathrm{~m}^{2}\). Also assume that the skillet has a constant temperature \(T_{s}=300^{\circ} \mathrm{C}\) and the drop has a temperature of \(100^{\circ} \mathrm{C}\). Water has density \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\), and the supporting layer has thermal conductivity \(k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} .\) (a) At what rate is energy conducted from the skillet to the drop through the drop's bottom surface? (b) If conduction is the primary way energy moves from the skillet to the drop, how long will the drop last?

(a) Two \(50 \mathrm{~g}\) ice cubes are dropped into \(200 \mathrm{~g}\) of water in a thermally insulated container. If the water is initially at \(25^{\circ} \mathrm{C}\), and the ice comes directly from a freezer at \(-15^{\circ} \mathrm{C}\), what is the final temperature at thermal equilibrium? (b) What is the final temperature if only one ice cube is used?

A flow calorimeter is a device used to measure the specific heat of a liquid. Energy is added as heat at a known rate to a stream of the liquid as it passes through the calorimeter at a known rate. Measurement of the resulting temperature difference between the inflow and the outflow points of the liquid stream enables us to compute the specific heat of the liquid. Suppose a liquid of density \(0.85 \mathrm{~g} / \mathrm{cm}^{3}\) flows through a calorimeter at the rate of \(8.0\) \(\mathrm{cm}^{3} / \mathrm{s}\). When energy is added at the rate of \(250 \mathrm{~W}\) by means of an electric heating coil, a temperature difference of \(15 \mathrm{C}^{\circ}\) is established in steady-state conditions between the inflow and the outflow points. What is the specific heat of the liquid?

(a) Calculate the rate at which body heat is conducted through the clothing of a skier in a steady-state process, given the following data: the body surface area is \(1.8 \mathrm{~m}^{2}\), and the clothing is \(1.0 \mathrm{~cm}\) thick; the skin surface temperature is \(33^{\circ} \mathrm{C}\) and the outer surface of the clothing is at \(1.0^{\circ} \mathrm{C}\); the thermal conductivity of the clothing is \(0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (b) If, after a fall, the skier's clothes became soaked with water of thermal conductivity \(0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), by how much is the rate of conduction multiplied?

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