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Chapter 18: Problem 50

A lab sample of gas is taken through cycle abca shown in the \(p-V\) diagram of Fig. 18 -43. The net work done is \(+1.2 \mathrm{~J}\). Along path \(a b,\) the change in the internal energy is \(+3.0 \mathrm{~J}\) and the magnitude of the work done is \(5.0 \mathrm{~J} .\) Along path \(c a,\) the energy transferred to the gas as heat is \(+2.5 \mathrm{~J}\). How much energy is transferred as heat along (a) path \(a b\) and (b) path \(b c ?\)

Short Answer

Expert verified
(a) Path \(ab: 8.0 \mathrm{~J}\); (b) Path \(bc: -9.3 \mathrm{~J}\).

Step by step solution

01

Understand the Problem

We are given a gas cycle represented in a \(p-V\) diagram (pressure vs. volume) and need to find the heat transfer for two segments of the cycle, \(a b\) and \(b c\). The given values are: net work done \(+1.2 \mathrm{~J}\), change in internal energy along \(a b\) is \(+3.0 \mathrm{~J}\), work done along \(a b\) is \(-5.0 \mathrm{~J}\), and heat transferred along \(c a\) is \(+2.5 \mathrm{~J}\).
02

Apply the First Law of Thermodynamics for Path ab

The First Law of Thermodynamics is stated as \(\Delta U = Q - W\), where \(\Delta U\) is the change in internal energy, \(Q\) is the heat added to the system, and \(W\) is the work done by the system. For path \(a b\), we know \(\Delta U = +3.0 \mathrm{~J}\) and \(W = -5.0 \mathrm{~J}\) (work done on the gas is positive). Substitute these into the equation to find \(Q_{ab}\): \(Q_{ab} = \Delta U + W = 3.0 + 5.0 = 8.0 \mathrm{~J}\).
03

Understand Energy Conservation in the Cycle

The entire cycle is a closed loop, so the sum of \(\Delta U\) over one complete cycle is zero. Therefore, for the cycle (\(abca\)), \(Q_{ab} + Q_{bc} + Q_{ca} = W_{net}\). We can use this to find the missing \(Q_{bc}\).
04

Calculate the Heat Transfer Qbc

We now use the energy balance over the whole cycle: \(Q_{ab} + Q_{bc} + Q_{ca} = 1.2 \mathrm{~J}\) (net work done). We found that \(Q_{ab} = 8.0 \mathrm{~J}\) and are given \(Q_{ca} = 2.5 \mathrm{~J}\). Substitute these into the cycle equation: \(8.0 + Q_{bc} + 2.5 = 1.2\). Solve for \(Q_{bc}\): \(Q_{bc} = 1.2 - 8.0 - 2.5 = -9.3 \mathrm{~J}\).
05

Summarize the Results

The heat transferred along path \(a b\) is \(8.0 \mathrm{~J}\), and along path \(b c\) is \(-9.3 \mathrm{~J}\). A negative heat transfer along \(b c\) indicates that heat was released from the gas in this segment.

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Key Concepts

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

p-V Diagram
A p-V diagram is a graph that plots the pressure (p) of a gas on the y-axis and its volume (V) on the x-axis. It is a valuable tool in thermodynamics for visualizing the behavior of gases and their transformations in cycles or processes.
The area under a curve in a p-V diagram represents the work done by or on a gas. In a closed loop, the net work done by the system is the area of the cycle traced out by the curve.
For this particular problem, we examine a cycle denoted as "abca," showing how the gas moves through specific paths (a to b, b to c, and c to a). The net work done during this cycle has been provided as +1.2 J. Understanding the work in each segment helps us apply the First Law of Thermodynamics to find values for heat transfer and changes in internal energy.
Internal Energy
Internal energy is a fundamental concept representing the total energy contained within a system. This includes both the kinetic energy due to the movement of particles and potential energy from forces between particles.
In thermodynamics, changes in internal energy (\(\Delta U\)) are important. They help determine how energy is transferred as heat and work. The First Law of Thermodynamics, which states \(\Delta U = Q - W\), connects these quantities together.
For path a to b in this cycle, the internal energy change is +3.0 J. This positive value means the system's internal energy increased, possibly due to an intake of heat or work being done on the gas. Calculating this change helps solve the cycle's energy balance and understand each segment's role in the overall cycle.
Heat Transfer
Heat transfer deals with the movement of thermal energy from one place to another or from one system to another, often occurring in the form of conduction, convection, or radiation.
In the context of this gas cycle, the First Law of Thermodynamics \(Q = \Delta U + W\) helps us determine the heat transfer values in each segment. For path ab, substituting known values, we find that \(Q_{ab}\) is 8.0 J. This indicates heat is added to the system during this section.
Conversely, for path bc, the problem's solution shows a negative heat transfer of -9.3 J. A negative value here tells us that the gas is losing heat during this segment. Overall, these calculations help piece together how energy flows differently throughout the cycle.

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Most popular questions from this chapter

A \(150 \mathrm{~g}\) copper bowl contains \(220 \mathrm{~g}\) of water, both at \(20.0^{\circ} \mathrm{C}\). A very hot \(300 \mathrm{~g}\) copper cylinder is dropped into the water, causing the water to boil, with \(5.00 \mathrm{~g}\) being converted to steam. The final temperature of the system is \(100^{\circ} \mathrm{C}\). Neglect energy transfers with the environment. (a) How much energy (in calories) is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature of the cylinder?

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 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?

What mass of butter, which has a usable energy content of \(6.0 \mathrm{Cal} / \mathrm{g}(=6000 \mathrm{cal} / \mathrm{g}),\) would be equivalent to the change in gravitational potential energy of a \(73.0 \mathrm{~kg}\) man who ascends from sea level to the top of Mt. Everest, at elevation \(8.84 \mathrm{~km}\) ? Assume that the average \(g\) for the ascent is \(9.80 \mathrm{~m} / \mathrm{s}^{2}\).

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