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For each of the situations described below, the object considered is undergoing some changes. Among the possible changes you should consider are: (Q) The object is absorbing or giving off thermal energy. (T) The object's temperature is changing. \(\left(E^{\text {int }}\right)\) The object's internal energy is changing. (W) The object is doing mechanical work or having work done on it. For each of the situations described below, identify which of the four changes are taking place and write as many of the letters \(\mathrm{Q}\), \(\mathrm{T}, E^{\text {int }}, \mathrm{W}\), (or none) as are appropriate. (a) A cylinder with a piston on top contains a compressed gas and is sitting on a thermal reservoir (a large iron block). After everything has come to thermal equilibrium, the piston is moved upward somewhat (very slowly). The object to be considered is the gas in the cylinder. (b) Consider the same cylinder as in part (a), but it is wrapped in styrofoam, a very good thermal insulator, instead of sitting on a thermal reservoir. The piston is pressed downward (again, very slowly), compressing the gas. The object to be considered is the

Short Answer

Expert verified
For (a), W. For (b), T, E^{int}, W.

Step by step solution

01

Understand the Problem Statement

You need to identify which changes are occurring (thermal energy change (Q), temperature change (T), internal energy change (E^{int}), or mechanical work (W)) in two different scenarios involving a gas in a cylinder with a piston.
02

Analyzing Situation (a)

Situation (a) describes a cylinder with a piston containing a compressed gas sitting on a thermal reservoir. The piston is moved upward very slowly after reaching thermal equilibrium.
03

Identify Changes in Situation (a)

Since the system is in thermal equilibrium, the temperature of the gas (T) does not change. Moving the piston upward slowly means work (W) is done by the gas as it expands. Since temperature (T) is constant, the internal energy (E^{int}) also remains constant. Thus, the changes involved are: W.
04

Analyzing Situation (b)

Situation (b) describes the same cylinder wrapped in styrofoam (a thermal insulator) where the piston is pressed downward, compressing the gas.
05

Identify Changes in Situation (b)

With the thermal insulator in place, thermal energy (Q) is not exchanged with the surroundings. Compressing the gas means work (W) is done on the gas. This compression increases the temperature (T) and the internal energy (E^{int}) of the gas. Therefore, the changes involved are: T, E^{int}, and W.

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Key Concepts

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

Thermal Energy Change
Thermal energy change occurs when heat is transferred into or out of an object. It's denoted by the symbol \(\text{Q}\). This energy transfer can happen through processes like conduction, convection, or radiation.

In thermodynamics, thermal energy change is crucial in understanding how heat affects a system. For instance, in situation (a), the gas cylinder is in contact with a thermal reservoir, which means it can exchange heat until it reaches thermal equilibrium.

In this equilibrium state, there is no net flow of thermal energy between the gas and the reservoir, so \(\text{Q}\) remains unchanged. In situation (b), the cylinder is insulated by styrofoam, preventing any thermal energy exchange with the surroundings. This means \(\text{Q}\) again stays constant.

Let's break it down with key points:
  • \textbf{Conduction:} Direct transfer of heat between objects in contact.
  • \textbf{Convection:} Transfer of heat through a fluid (liquid or gas).
  • \textbf{Radiation:} Transfer of heat through electromagnetic waves.
Temperature Change
Temperature change, denoted by \(\text{T}\), occurs when there is a variation in the kinetic energy of the particles within a substance. The temperature provides a measure of the average kinetic energy of these particles.

In situation (a), since the gas is in thermal equilibrium with the reservoir, its temperature doesn't change, so \(\text{T}\) remains constant. In contrast, in situation (b), compressing the gas increases its temperature due to the work done on it. Styrofoam insulation ensures that the system doesn't lose or gain heat, causing the temperature to rise.

Key points on temperature changes:
  • \textbf{Increase:} When energy is added to a system, raising the kinetic energy of particles.
  • \textbf{Decrease:} When energy is removed, lowering the kinetic energy of particles.
Internal Energy Change
Internal energy change, symbolized as \(\text{E}^{\text{int}}\), involves alterations in the total energy contained within a system due to changes in kinetic and potential energy at a microscopic level. This energy is the sum of all molecular energies.

In situation (a), with the gas at thermal equilibrium and temperature constant, there's no change in internal energy, so \(\text{E}^{\text{int}}\) doesn't change. However, in situation (b), compressing the gas not only does mechanical work on it but also increases its internal energy. This is because the molecules move faster and more energetically when compressed.

Important aspects of internal energy changes:
  • \textbf{Energy Added:} Raises internal energy.
  • \textbf{Energy Removed:} Lowers internal energy.
  • \textbf{Work Done:} Can change internal energy by compressing or expanding gas.
Mechanical Work
Mechanical work, designated as \(\text{W}\), occurs when a force acts on an object causing displacement. In thermodynamics, it often involves the work done by or on gases within a system.

For situation (a), moving the piston upward means the gas is doing work as it expands against the piston, represented by \(\text{W}\). This expansion is done slowly, allowing the system to remain in thermal equilibrium without changing its temperature or internal energy.

In situation (b), pressing the piston downward means mechanical work is done on the gas. This compression increases both the gas's temperature and internal energy. Insulation ensures no thermal energy is exchanged with the environment, but work done on the gas increases its internal energy.

Key points on mechanical work:
  • \textbf{Positive Work:} When the system does work on its surroundings (expansion).
  • \textbf{Negative Work:} When work is done on the system (compression).
  • \textbf{Energy Transfer:} Work transfers energy into or out of the system.

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Most popular questions from this chapter

A tank of water has been outdoors in cold weather, and a slab of ice \(5.0 \mathrm{~cm}\) thick has formed on its surface (Fig. 19-42). The air above the ice is at \(-10^{\circ} \mathrm{C}\). Calculate the rate of formation of ice (in centimeters per hour) on the ice slab. Take the thermal conductivity and density of ice to be \(0.0040 \mathrm{cal} / \mathrm{s} \cdot \mathrm{cm} \cdot \mathrm{C}^{\circ}\) and \(0.92 \mathrm{~g} / \mathrm{cm}^{3}\). Assume that energy is not transferred through the walls or bottom of the tank.

(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) How would the answer to (a) change if, after a fall, the skier's clothes became soaked with water of thermal conductivity \(0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} ?\)

A cylindrical copper rod of length \(1.2 \mathrm{~m}\) and cross-sectional area \(4.8 \mathrm{~cm}^{2}\) is insulated to prevent thermal energy from being transferred through its surface. The ends are maintained at a temperature difference of \(100^{\circ} \mathrm{C}\) by having one end in a water-ice mixture and the other in boiling water and steam. (a) Find the rate at which thermal energy is conducted along the rod. (b) Find the rate at which ice melts at the cold end.

Consider that 200 J of work is done on a system and \(70.0\) cal of thermal energy is trans-ferred out of the system. In the sense of the first law of thermodynamics, what are the values (including algebraic signs) of (a) \(W,(\mathrm{~b})\) \(Q\), and \((\mathrm{c}) \Delta E^{\mathrm{int}} ?\)

A chef, on finding his stove out of order, decides to boil the water for his wife's coffee by shaking it in a thermos flask. Suppose that he uses tap water at \(15^{\circ} \mathrm{C}\) and that the water falls \(30 \mathrm{~cm}\) each shake, the chef making 30 shakes each minute. Neglecting any transfer of thermal energy out of the flask, how long must he shake the flask for the water to reach \(100^{\circ} \mathrm{C}\) ?

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