1. Two systems \(X\) and \(Y\) are placed in thermal communication and allowed sufficient time to interact. Which observation identifies that they have attained thermal equilibrium with each other?
ⓐ. Their pressures become equal even though their temperatures remain different.
ⓑ. Their temperatures become equal and there is no net heat transfer between them.
ⓒ. Their internal energies become equal while their temperatures may remain different.
ⓓ. Their masses and volumes become equal and further heat transfer becomes impossible.
Correct Answer: Their temperatures become equal and there is no net heat transfer between them.
Explanation: Thermal equilibrium between two systems is characterised by equality of their temperatures. If the systems can exchange heat, an initial temperature difference produces a net transfer of energy from the hotter system toward the colder one. This temperature-driven transfer changes their thermal states as they approach equilibrium. At equilibrium, the temperature difference has vanished, so there is no net heat flow between the systems. Equality of pressure is not required because thermal equilibrium is distinct from mechanical equilibrium. Equality of internal energy is also unnecessary because two different systems can possess different internal energies while having the same temperature. Their masses and volumes likewise need not be equal. Therefore, equal temperature together with absence of net temperature-driven heat transfer is the correct criterion.
2. Consider the following statements about two systems that are in mutual thermal equilibrium.
Statement I: The systems may have different masses, volumes and internal energies.
Statement II: The systems must have equal pressure and equal volume in addition to equal temperature.
ⓐ. Both Statement I and Statement II are correct.
ⓑ. Both Statement I and Statement II are incorrect.
ⓒ. Statement I is incorrect but Statement II is correct.
ⓓ. Statement I is correct but Statement II is incorrect.
Correct Answer: Statement I is correct but Statement II is incorrect.
Explanation: Statement I is correct because thermal equilibrium imposes a condition on temperature, not on every macroscopic property of the systems. Two bodies made of different substances can have equal temperature even though their masses and internal energies are very different. Their volumes may also be different. Thermal equilibrium means that, when thermal communication is permitted, there is no net heat transfer driven by a temperature difference. Statement II is incorrect because equal pressure belongs to a mechanical-equilibrium consideration and is not required for thermal equilibrium. Equal volume has no such requirement either. A small object and a large object can readily be in thermal equilibrium if their temperatures are equal. Thus the decisive variable here is temperature rather than pressure, volume, mass or total internal energy. Hence only Statement I is correct.
3. Use the arrangement described below.
System \(P\) is initially at \(350\,\text{K}\) and system \(Q\) is initially at \(300\,\text{K}\). They are separated by an adiabatic wall and the pair is thermally isolated from everything else.
Which statement best describes the subsequent thermal behaviour?
ⓐ. Heat flows from \(P\) to \(Q\) until both reach \(325\,\text{K}\).
ⓑ. Heat flows from \(P\) to \(Q\) until their internal energies become equal.
ⓒ. Their temperatures necessarily become equal because the pair is isolated.
ⓓ. Their temperatures can remain unequal because the separating wall prevents heat transfer.
Correct Answer: Their temperatures can remain unequal because the separating wall prevents heat transfer.
Explanation: An adiabatic wall is an insulating thermal boundary, so it prevents heat transfer through that wall. The fact that \(P\) is hotter than \(Q\) provides a tendency for heat to flow from \(P\) toward \(Q\), but the required thermal path is blocked. Therefore, the temperature difference alone does not guarantee that the two systems will approach mutual thermal equilibrium. Thermal isolation of the whole pair merely prevents energy exchange with the external surroundings; it does not turn the internal adiabatic wall into a conducting wall. The value \(325\,\text{K}\) cannot be inferred merely by averaging the two temperatures. Such a calculation would additionally require appropriate thermal-property information. Equality of internal energies is not the criterion for thermal equilibrium in any case. Also, the word adiabatic must not be confused with constant temperature. Here the defining fact is simply that no heat can cross the separating boundary, so the unequal temperatures may persist.
4. Two bodies \(A\) and \(B\), initially at \(360\,\text{K}\) and \(300\,\text{K}\) respectively, are first separated by an adiabatic wall. The wall is then replaced by a conducting wall while the pair remains isolated from the external surroundings. Which sequence is physically correct?
ⓐ. Net heat flows from \(A\) to \(B\), the temperature difference decreases, and net heat transfer stops when their temperatures become equal.
ⓑ. Net heat flows from \(B\) to \(A\), the temperature difference increases, and the process stops when their internal energies become equal.
ⓒ. No heat flows because the pair as a whole is isolated from the surroundings.
ⓓ. Both temperatures remain fixed because replacing a wall cannot change the thermal state of either body.
Correct Answer: Net heat flows from \(A\) to \(B\), the temperature difference decreases, and net heat transfer stops when their temperatures become equal.
Explanation: Initially the adiabatic wall blocks heat transfer between the two bodies, so the temperature difference can persist. Replacing that wall by a conducting wall creates thermal communication between \(A\) and \(B\). Since \(T_A=360\,\text{K}\) and \(T_B=300\,\text{K}\), the spontaneous net heat transfer is from the higher-temperature body \(A\) toward the lower-temperature body \(B\). As energy is exchanged, the temperature difference between them decreases. The process continues until a common temperature is reached. At that stage there is no temperature difference to sustain a net heat flow, so the bodies are in thermal equilibrium. External isolation does not prevent this internal exchange; it only prevents heat exchange between the two-body composite and its surroundings. Equal internal energies are neither required nor generally expected. Thus the correct sequence is hotter-to-colder transfer followed by equality of temperature and cessation of net heat flow.
5. Consider the following statements concerning two systems in thermal equilibrium.
Statement I: There is no net heat transfer between the systems.
Statement II: All microscopic energy exchanges between the systems must cease completely.
Statement III: The systems can have unequal internal energies.
Statement IV: Different masses prevent the systems from being in thermal equilibrium.
ⓐ. I and II only
ⓑ. I, III and IV only
ⓒ. II and IV only
ⓓ. I and III only
Correct Answer: I and III only
Explanation: Statement I is correct because thermal equilibrium corresponds to zero net heat transfer between systems in thermal communication. Statement II is incorrect because the macroscopic condition of zero net transfer does not require microscopic exchanges to stop altogether. Microscopic exchanges can occur in both directions while their average effects balance so that the net transfer is zero. Statement III is correct because internal energy depends on the system and its state, and equal temperature does not imply equal total internal energy. For example, systems having different masses or compositions may have very different internal energies at the same temperature. Statement IV is incorrect because equality of mass is not a condition for thermal equilibrium. A small body and a large body can be at exactly the same temperature and exhibit no net heat flow when placed in thermal contact. Thermal equilibrium must therefore not be confused with equality of total energy or amount of matter. Hence Statements I and III alone are correct.
6. Use the graph description below.
The vertical axis represents temperature and the horizontal axis represents time. Curve \(X\) starts at \(360\,\text{K}\) and decreases smoothly. Curve \(Y\) starts at \(300\,\text{K}\) and increases smoothly. The curves meet at a common temperature and thereafter remain coincident and horizontal.
Which interpretation is consistent with the graph?
ⓐ. \(Y\) initially transfers heat to \(X\), and equality of temperature later reverses the heat flow.
ⓑ. The two bodies must have equal internal energies when the curves meet.
ⓒ. The graph proves that the two bodies have equal masses and heat capacities.
ⓓ. \(X\) initially transfers net heat to \(Y\), and the common horizontal portion represents thermal equilibrium.
Correct Answer: \(X\) initially transfers net heat to \(Y\), and the common horizontal portion represents thermal equilibrium.
Explanation: At the start, curve \(X\) represents the higher-temperature body because \(360\,\text{K}\) exceeds \(300\,\text{K}\). Therefore, if the two bodies are in conducting thermal contact, net heat transfer is from \(X\) toward \(Y\). This is consistent with \(X\) cooling while \(Y\) warms. The narrowing separation between the curves represents a decreasing temperature difference as equilibrium is approached. When the curves meet, both systems have the same temperature. The later horizontal coincidence indicates that this equality persists and there is no further net temperature-driven heat transfer. The graph does not imply equality of total internal energies because that quantity also depends on the nature and amount of each system. It also does not establish equal masses or equal heat capacities. The relevant graphical feature is convergence to a common temperature, not equality of unrelated physical properties. Thus the graph represents approach to and attainment of thermal equilibrium.
7. A large body \(L\) at \(300\,\text{K}\) has greater total internal energy than a much smaller body \(S\) at \(340\,\text{K}\). The two are brought into conducting thermal contact. Which statement is correct immediately after contact?
ⓐ. Net heat flows from \(S\) to \(L\) because \(S\) has the higher temperature.
ⓑ. Net heat flows from \(L\) to \(S\) because \(L\) has greater internal energy.
ⓒ. No net heat flows because the bodies have opposing temperature and energy differences.
ⓓ. No net heat flows until their total internal energies become equal.
Correct Answer: Net heat flows from \(S\) to \(L\) because \(S\) has the higher temperature.
Explanation: The direction of spontaneous heat transfer is determined by temperature difference, not by which body contains more total internal energy. Here \(S\) is at \(340\,\text{K}\), whereas \(L\) is at \(300\,\text{K}\). Therefore, the net heat transfer begins from \(S\), the hotter body, toward \(L\), the colder body. The fact that \(L\) has a larger total internal energy does not reverse this direction. Internal energy is an extensive property and can be larger simply because a system contains much more matter. Temperature is the relevant thermal variable for deciding the direction of heat exchange between bodies placed in thermal contact. As the interaction proceeds, the temperature difference decreases. Net heat transfer ceases when the temperatures become equal. Even at that final thermal-equilibrium state, the internal energies need not be equal. This is why comparing total internal energies instead of temperatures is an important but incorrect shortcut.
8. Study the three cases below.
| Case | Left system temperature | Right system temperature | Separating wall |
|---|
| I | \(320\,\text{K}\) | \(320\,\text{K}\) | Conducting |
| II | \(340\,\text{K}\) | \(300\,\text{K}\) | Adiabatic |
| III | \(350\,\text{K}\) | \(300\,\text{K}\) | Conducting |
Assuming each pair is otherwise thermally isolated, in which case will spontaneous temperature-driven thermal evolution occur through the separating wall?
ⓐ. Case I only
ⓑ. Cases I and II only
ⓒ. Cases II and III only
ⓓ. Case III only
Correct Answer: Case III only
Explanation: Case I has a conducting wall, but both systems already have the same temperature of \(320\,\text{K}\). Therefore, there is no temperature difference to produce net heat transfer, and the pair is already in thermal equilibrium. Case II has a temperature difference, but the adiabatic wall blocks heat transfer through the boundary. Consequently, that temperature difference does not by itself cause the two systems to evolve toward mutual thermal equilibrium through that wall. Case III combines both requirements for spontaneous thermal evolution: the temperatures are unequal and the separating wall permits heat transfer. Net heat therefore flows from the \(350\,\text{K}\) system toward the \(300\,\text{K}\) system. Their temperature difference then decreases as they approach equilibrium. The comparison shows that a temperature difference provides the driving condition, while a conducting boundary provides the permitted thermal path. Either absence of a temperature difference or blockage by an adiabatic boundary prevents net temperature-driven transfer through the stated wall. Hence only Case III evolves in the required way.
9. Use the two-stage arrangement described below.
System \(A\) is at \(360\,\text{K}\). Systems \(B\) and \(C\) are initially in conducting contact while \(A\) is separated from \(B\) by an adiabatic wall. After sufficient time, \(B\) and \(C\) are observed to be in thermal equilibrium at \(300\,\text{K}\). The \(B-C\) wall is then made adiabatic and, at the same instant, the \(A-B\) wall is replaced by a conducting wall.
What happens immediately after the second-stage change?
ⓐ. Net heat flows from \(B\) to \(A\) because \(B\) has just completed a thermal-equilibrium process.
ⓑ. Heat must first flow from \(C\) to \(B\) before any transfer between \(A\) and \(B\) can occur.
ⓒ. No net heat flows across \(A-B\) because \(B\) was previously in thermal equilibrium with \(C\).
ⓓ. Net heat flows from \(A\) to \(B\) because the newly communicating systems are at unequal temperatures.
Correct Answer: Net heat flows from \(A\) to \(B\) because the newly communicating systems are at unequal temperatures.
Explanation: During the first stage, \(B\) and \(C\) are allowed to reach a common temperature of \(300\,\text{K}\). At the start of the second stage, the \(B-C\) boundary is made adiabatic, so the previous thermal connection between those two systems is removed. At the same time, \(A\) at \(360\,\text{K}\) is placed in conducting thermal contact with \(B\) at \(300\,\text{K}\). The relevant comparison is therefore the temperature of the systems connected by the newly conducting wall. Since \(A\) is hotter, net heat begins to flow from \(A\) toward \(B\). The fact that \(B\) had previously been in equilibrium with \(C\) does not freeze \(B\)'s thermal state once its boundary conditions are changed. Thermal equilibrium is a condition applying to a specified situation, not a permanent label attached to a body. Since \(C\) is now separated from \(B\) by an adiabatic wall, no heat from \(C\) is required for the \(A-B\) interaction to begin. The second-stage behaviour is thus determined by the new wall configuration together with the new pairwise temperature difference.
10. Bodies \(P\) and \(Q\) are placed in conducting contact and allowed to reach thermal equilibrium. They are then thermally separated from each other. Next, \(Q\) is brought into conducting contact with a colder body \(R\) and the \(Q-R\) pair is allowed to reach thermal equilibrium while \(P\) remains isolated. Which statement is necessarily correct at the end?
ⓐ. \(P\) and \(Q\) must still be in thermal equilibrium because they once had equal temperatures.
ⓑ. All three bodies must have the same temperature because thermal equilibrium cannot be lost.
ⓒ. \(Q\) and \(R\) are in thermal equilibrium, while \(P\) and \(Q\) need not remain so.
ⓓ. \(P\) must become colder even though it remains thermally isolated.
Correct Answer: \(Q\) and \(R\) are in thermal equilibrium, while \(P\) and \(Q\) need not remain so.
Explanation: After the first interaction, \(P\) and \(Q\) have the same temperature and are in mutual thermal equilibrium. They are then thermally separated, so subsequent changes in \(Q\) are not automatically transmitted to \(P\). When \(Q\) is placed in conducting contact with the colder body \(R\), net heat flows from \(Q\) toward \(R\). Consequently, the temperature of \(Q\) changes while \(P\) remains isolated and retains its state under the stated conditions. The \(Q-R\) exchange continues until those two bodies reach a common temperature. At that time, \(Q\) and \(R\) are certainly in thermal equilibrium. However, because \(Q\)'s temperature has changed after separation from \(P\), the previous equality of \(P\) and \(Q\) need not survive. Thermal equilibrium is therefore not a permanent historical relationship between two bodies. It depends on their current thermal states and boundary conditions. This sequential-contact situation distinguishes a present equilibrium condition from the fact that two bodies happened to be in equilibrium at an earlier time.
11. Assertion: If two systems separated by an adiabatic wall show no temperature change, and replacing that wall by a conducting wall still produces no temperature-driven thermal evolution, the observation is consistent with the two systems already having equal temperatures.
Reason: A conducting wall permits heat transfer, but equal temperatures provide no temperature difference for a net heat flow.
ⓐ. Both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.
ⓑ. Both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
ⓒ. Assertion is correct, but Reason is incorrect.
ⓓ. Assertion is incorrect, but Reason is correct.
Correct Answer: Both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.
Explanation: The Assertion is correct under the stated thermal conditions. An adiabatic wall can conceal a temperature difference because it prevents heat transfer, so the absence of evolution while that wall is present does not by itself prove thermal equilibrium. The crucial observation comes after the boundary is replaced by a conducting wall. If the systems had different temperatures, the conducting boundary would permit a spontaneous net transfer of heat from the hotter system toward the colder one. The observation that no such temperature-driven evolution occurs is therefore consistent with the absence of a temperature difference. The Reason is also correct because a conducting wall provides the route for heat transfer but does not itself create the driving temperature difference. When temperatures are equal, opposing microscopic exchanges produce no net temperature-driven transfer. This directly explains why insertion of a conducting wall causes no macroscopic thermal change in the stated case. The Reason therefore supplies the physical basis of the Assertion. Hence both are correct and the Reason correctly explains the Assertion.
12. Two thermally isolated bodies \(X\) and \(Y\) begin at \(360\,\text{K}\) and \(300\,\text{K}\), respectively. They are placed in conducting contact at \(t=0\). Before thermal equilibrium is reached, at time \(t=t_1\), the conducting boundary is replaced by an adiabatic boundary. Use the graph description below.
The vertical axis is temperature and the horizontal axis is time. The required graph must describe the temperatures of both bodies before and after the boundary change at \(t=t_1\).
Which behaviour is consistent with the stated process?
ⓐ. Both temperatures remain fixed from \(t=0\) because the complete two-body system is isolated.
ⓑ. \(X\) cools and \(Y\) warms until \(t_1\), after which both continue changing until their temperatures become equal.
ⓒ. \(X\) warms and \(Y\) cools until \(t_1\), after which their temperatures suddenly become equal.
ⓓ. \(X\) cools and \(Y\) warms until \(t_1\), after which their then-current unequal temperatures remain unchanged under the stated isolation.
Correct Answer: \(X\) cools and \(Y\) warms until \(t_1\), after which their then-current unequal temperatures remain unchanged under the stated isolation.
Explanation: From \(t=0\) to \(t_1\), the two bodies are connected by a conducting boundary and have unequal temperatures. Net heat therefore flows from the hotter body \(X\) toward the colder body \(Y\). Consequently, \(T_X\) decreases while \(T_Y\) increases, and the separation between the two temperature curves becomes smaller. The problem explicitly states that the conducting wall is removed before equilibrium has been reached, so the temperatures are still unequal at \(t_1\). At that instant, the replacement adiabatic boundary blocks further heat transfer between the bodies. Because the pair is also thermally isolated from external surroundings, there is no remaining thermal path by which the two temperatures can continue approaching each other. Thus each curve becomes horizontal at its own value after \(t_1\). There is no physical reason for an instantaneous jump to a common temperature when the insulating wall is introduced. The graph therefore demonstrates how changing the boundary condition can interrupt an ongoing approach to thermal equilibrium.
13. Three bodies \(A\), \(B\) and \(C\) have initial temperatures \(360\,\text{K}\), \(320\,\text{K}\) and \(280\,\text{K}\), respectively. They have finite positive heat capacities and undergo no phase change. In Stage 1, only \(A\) and \(B\) are placed in conducting contact and are allowed to reach thermal equilibrium. They are then separated by an adiabatic wall. In Stage 2, \(B\) is placed in conducting contact with \(C\). Which sequence gives the directions of net heat transfer in the two stages?
ⓐ. Stage 1: \(B\rightarrow A\); Stage 2: \(C\rightarrow B\)
ⓑ. Stage 1: \(A\rightarrow B\); Stage 2: \(B\rightarrow C\)
ⓒ. Stage 1: \(A\rightarrow B\); Stage 2: \(C\rightarrow B\)
ⓓ. Stage 1: no net transfer; Stage 2: \(B\rightarrow C\)
Correct Answer: Stage 1: \(A\rightarrow B\); Stage 2: \(B\rightarrow C\)
Explanation: In Stage 1, \(A\) begins at \(360\,\text{K}\) and \(B\) at \(320\,\text{K}\), so net heat flows from \(A\) to \(B\). Because both bodies have finite positive heat capacities and no phase change occurs, their common equilibrium temperature lies between their initial temperatures. Therefore, at the end of Stage 1, the temperature of \(B\) is greater than its original \(320\,\text{K}\) but less than \(360\,\text{K}\). The \(A-B\) connection is then replaced by an adiabatic separation, so the subsequent behaviour of \(B\) is considered with \(C\). Body \(C\) is at only \(280\,\text{K}\). Thus, when \(B\) is connected conductively to \(C\), \(B\) is certainly hotter than \(C\). Net heat in Stage 2 must therefore flow from \(B\) toward \(C\). Notice that no numerical equilibrium-temperature calculation is needed; only its qualitative position relative to the initial temperatures is required. The problem consequently combines heat-flow direction, approach to equilibrium and a changed thermal boundary. The correct sequence is \(A\rightarrow B\) followed by \(B\rightarrow C\).
14. Four systems have the following initial arrangement. Systems \(A\) and \(B\) are both at \(310\,\text{K}\) and are separated by a conducting wall. Systems \(C\) and \(D\) are both at \(350\,\text{K}\) and are separated by another conducting wall. The \(B-C\) boundary is adiabatic. Consider the statements:
Statement I: \(A\) and \(B\) are in thermal equilibrium with each other.
Statement II: \(C\) and \(D\) are in thermal equilibrium with each other.
Statement III: \(B\) and \(C\) are in thermal equilibrium because no heat crosses their adiabatic boundary.
Statement IV: If the \(B-C\) boundary is replaced by a conducting wall, net heat initially flows from \(C\) toward \(B\).
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I, II and IV only
ⓓ. I, III and IV only
Correct Answer: I, II and IV only
Explanation: Statement I is correct because \(A\) and \(B\) have equal temperatures and are in conducting thermal communication. There is therefore no net temperature-driven heat flow between them. Statement II is correct for the same reason: \(C\) and \(D\) are both at \(350\,\text{K}\) and their conducting connection produces no net heat transfer. Statement III is incorrect because absence of heat transfer across an adiabatic wall does not establish thermal equilibrium. Here \(B\) is at \(310\,\text{K}\) while \(C\) is at \(350\,\text{K}\), so they are not at the same temperature. The insulating wall merely prevents their temperature difference from producing heat exchange through that boundary. Statement IV is correct because replacing the adiabatic wall with a conducting wall permits the existing temperature difference to act. Since \(C\) is hotter than \(B\), the initial net heat flow is from \(C\) toward \(B\). This contrast highlights the essential distinction between “no heat transfer because temperatures are equal” and “no heat transfer because the thermal path is blocked.” Therefore Statements I, II and IV are correct.
15. A large system \(L\) at \(300\,\text{K}\) possesses greater total internal energy than a smaller system \(S\) at \(360\,\text{K}\). Initially they are separated by an adiabatic wall. Later that wall is replaced by a conducting wall, and the systems are allowed to reach thermal equilibrium. Which description is correct?
ⓐ. No energy transfer occurs after the wall change because the system with greater internal energy is already \(L\).
ⓑ. Net heat first flows from \(S\) to \(L\); at equilibrium their temperatures are equal, but their internal energies need not be equal.
ⓒ. Net heat first flows from \(L\) to \(S\); it stops when the total internal energies become equal.
ⓓ. Net heat first flows from \(S\) to \(L\); it stops when both temperature and total internal energy become equal.
Correct Answer: Net heat first flows from \(S\) to \(L\); at equilibrium their temperatures are equal, but their internal energies need not be equal.
Explanation: While the adiabatic wall is present, heat transfer between \(L\) and \(S\) is blocked even though their temperatures differ. Once that wall is replaced by a conducting one, the relevant driving variable is temperature. System \(S\) is at \(360\,\text{K}\), so it is hotter than \(L\) at \(300\,\text{K}\). Net heat therefore flows from \(S\) toward \(L\). The larger total internal energy of \(L\) does not reverse the direction because internal energy and temperature are not interchangeable quantities. As heat is exchanged, the temperature difference decreases. Thermal equilibrium is reached when both systems have a common temperature and the net heat transfer becomes zero. There is no requirement that their total internal energies become equal at that point. Different masses, materials or other state characteristics can produce unequal internal energies even at a common temperature. The initial adiabatic stage and the later conducting stage also show that both temperature difference and boundary type matter. Thus the correct conclusion separates the equilibrium temperature criterion from the misconception of equal internal energy.
16. Use the multi-stage arrangement described below.
Four finite bodies \(A,B,C,D\) have positive heat capacities and undergo no phase change. Initially \(A\) and \(B\) are in conducting contact and thermally isolated from \(C\) and \(D\). After sufficient time, \(A\) and \(B\) are observed at a common temperature of \(330\,\text{K}\). Independently, \(C\) and \(D\) are in conducting contact at \(340\,\text{K}\). Next, the \(A-B\) and \(C-D\) boundaries are both made adiabatic, while \(B\) and \(C\) are connected by a conducting wall. The system is then left until \(B\) and \(C\) reach thermal equilibrium.
Which statement is necessarily correct after the final stage?
ⓐ. All four bodies must be in thermal equilibrium because each body belonged to an equilibrium pair at some stage.
ⓑ. \(A\), \(B\) and \(C\) must have the same temperature, while \(D\) remains at \(340\,\text{K}\).
ⓒ. \(A\) and \(D\) must acquire equal temperatures because they were insulated during the final interaction.
ⓓ. \(B\) and \(C\) are in thermal equilibrium with each other, while equality of their final temperature with that of \(A\) or \(D\) is not implied.
Correct Answer: \(B\) and \(C\) are in thermal equilibrium with each other, while equality of their final temperature with that of \(A\) or \(D\) is not implied.
Explanation: Before the final rearrangement, \(A\) and \(B\) are both at \(330\,\text{K}\), whereas \(C\) and \(D\) are both at \(340\,\text{K}\). When the original conducting connections are replaced by adiabatic boundaries, \(A\) is thermally separated from \(B\) and \(D\) is thermally separated from \(C\). The new conducting connection is only between \(B\) and \(C\). Since \(C\) begins that stage at the higher temperature, net heat initially flows from \(C\) toward \(B\). Their temperatures approach each other until \(B\) and \(C\) reach a common equilibrium temperature. Because both have finite positive heat capacities and no phase change occurs, this common value lies between \(330\,\text{K}\) and \(340\,\text{K}\). Body \(A\), being insulated during this stage, remains at \(330\,\text{K}\) under the stated conditions. Body \(D\) likewise remains at \(340\,\text{K}\). Therefore the new common temperature of \(B\) and \(C\) is not forced to equal either isolated body's temperature. The fact that different pairs were in equilibrium at earlier times does not make all four systems permanently mutually equilibrated. Thermal equilibrium must always be assessed from the current temperatures and current thermal connections. Hence only the \(B-C\) equilibrium is necessarily established at the end.
17. Three independent pairs of systems are observed under the conditions shown below. Each pair is isolated from external thermal influence and is allowed sufficient time after the stated wall is established.
| Pair | Temperatures when observation begins | Separating wall | Observation |
|---|
| I | \(315\,\text{K},315\,\text{K}\) | Conducting | No macroscopic thermal change |
| II | \(340\,\text{K},300\,\text{K}\) | Adiabatic | No macroscopic thermal change |
| III | Not disclosed | Conducting | No macroscopic thermal change |
Which conclusion is justified?
ⓐ. All three pairs must be in thermal equilibrium because none shows a thermal change.
ⓑ. Only Pair II must be in thermal equilibrium because an adiabatic wall guarantees equilibrium.
ⓒ. Pairs I and III can be identified as thermally equilibrated, whereas the observation for Pair II does not establish thermal equilibrium.
ⓓ. Only Pair I can be identified as thermally equilibrated because the temperatures of Pair III are not numerically given.
Correct Answer: Pairs I and III can be identified as thermally equilibrated, whereas the observation for Pair II does not establish thermal equilibrium.
Explanation: Pair I has equal temperatures and a conducting boundary, so there is no temperature difference capable of producing net heat flow. Its lack of thermal evolution is therefore consistent with thermal equilibrium. Pair II is fundamentally different because its two temperatures are unequal. The absence of thermal change there results from the adiabatic wall blocking heat transfer, not from equality of temperature. Hence “no observed heat transfer” does not by itself prove thermal equilibrium unless the nature of the boundary is considered. Pair III has a conducting wall, so a temperature difference would have produced spontaneous thermal evolution under the stated isolated conditions. Since no such change occurs even after sufficient time, the consistent inference is that its two systems already have equal temperatures. Numerical temperature values are unnecessary for this reverse inference. These conditions therefore separate two physically different reasons for observing no net thermal evolution: equality of temperature and absence of a conducting path. Only the former establishes mutual thermal equilibrium.
18. Use the arrangement described below.
Three finite systems \(A,B,C\) are isolated from the surroundings. The \(A-B\) and \(B-C\) boundaries are conducting, while the direct \(A-C\) boundary is adiabatic. The systems are left undisturbed for sufficient time until no further macroscopic thermal changes occur.
Which conclusion must then hold?
ⓐ. \(A\) and \(C\) may necessarily remain at different temperatures because their direct common wall is adiabatic.
ⓑ. \(T_A=T_B=T_C\), although pressure, volume and internal energy need not be equal for the three systems.
ⓒ. Only \(T_A=T_B\) is required because thermal information cannot pass through system \(B\).
ⓓ. The three internal energies must become equal because no net heat transfer remains anywhere in the arrangement.
Correct Answer: \(T_A=T_B=T_C\), although pressure, volume and internal energy need not be equal for the three systems.
Explanation: The conducting \(A-B\) boundary allows heat exchange until \(A\) and \(B\) have equal temperatures. Independently, the conducting \(B-C\) boundary allows heat exchange until \(B\) and \(C\) have equal temperatures. At the final stationary thermal state,
\[
T_A=T_B
\]
and
\[
T_B=T_C.
\]
Combining these relations gives
\[
T_A=T_B=T_C.
\]
The direct adiabatic \(A-C\) wall does not prevent this result because \(A\) and \(C\) are thermally linked indirectly through \(B\). An adiabatic boundary blocks heat transfer across that particular boundary; it does not force systems on its two sides to have different temperatures. Equal temperature also does not require equal pressures, volumes or internal energies. Thus the network of conducting contacts establishes a common thermal-equilibrium temperature even though one direct boundary remains insulating.
19. Three bodies have initial temperatures \(T_A=300\,\text{K}\), \(T_B=340\,\text{K}\) and \(T_C=350\,\text{K}\). Initially only \(A\) and \(B\) are in conducting contact. Before they reach equilibrium, \(B\) is observed at \(325\,\text{K}\). At that instant the \(A-B\) boundary is made adiabatic and \(B\) is placed in conducting contact with \(C\). The whole arrangement is isolated from the surroundings. Which description of body \(B\) is correct?
ⓐ. \(B\) receives heat in both stages because it always has more internal energy than \(A\).
ⓑ. \(B\) loses heat in both stages because it started above \(300\,\text{K}\).
ⓒ. \(B\) loses heat in the first stage and then remains thermally unchanged in the second stage.
ⓓ. \(B\) loses net heat in the first stage but receives net heat in the second stage.
Correct Answer: \(B\) loses net heat in the first stage but receives net heat in the second stage.
Explanation: During the first stage, \(B\) is initially at \(340\,\text{K}\) while \(A\) is at \(300\,\text{K}\). Therefore net heat flows from \(B\) toward \(A\), and the fall of \(B\)'s temperature to \(325\,\text{K}\) is consistent with this direction. The first contact is then interrupted by making the \(A-B\) wall adiabatic. At that moment, \(B\) is connected instead to \(C\), which is at \(350\,\text{K}\). The relevant comparison has therefore changed from \(T_B\gt T_A\) to \(T_B\lt T_C\). Net heat in the second stage consequently flows from \(C\) toward \(B\). Thus the same body changes from being the donor of net heat to being the receiver because its thermal neighbour has changed. The direction is decided separately at each stage from the temperatures of the systems that are actually in conducting contact. A body's initial temperature or total internal energy does not permanently determine the direction of all later heat exchanges.
20. Two isolated bodies \(P\) and \(Q\) begin at \(350\,\text{K}\) and \(300\,\text{K}\), respectively. From \(t=0\) to \(t=t_1\) they are separated by a conducting wall. Before equilibrium is reached, the wall is made adiabatic from \(t_1\) to \(t_2\). At \(t=t_2\), the conducting wall is restored and remains so thereafter. Use the graph description below.
The vertical axis is temperature and the horizontal axis is time. The required pair of temperature curves must represent the complete three-stage wall sequence.
Which qualitative behaviour is physically consistent?
ⓐ. The temperatures approach each other until \(t_1\), remain fixed at unequal values until \(t_2\), and then resume approaching a common value.
ⓑ. The temperatures approach each other until \(t_1\), become equal instantly when the wall is made adiabatic, and remain equal thereafter.
ⓒ. The temperatures remain unchanged until \(t_1\), approach each other only while the wall is adiabatic, and become fixed when conduction is restored.
ⓓ. The temperatures approach each other continuously through all three stages because the earlier heat flow cannot be interrupted by changing the wall.
Correct Answer: The temperatures approach each other until \(t_1\), remain fixed at unequal values until \(t_2\), and then resume approaching a common value.
Explanation: During the first stage, \(P\) is hotter than \(Q\) and the wall is conducting. Therefore net heat flows from \(P\) to \(Q\), so \(T_P\) falls while \(T_Q\) rises. The problem states that equilibrium has not yet been reached at \(t_1\), so the temperatures are still unequal when the wall is changed. Making the boundary adiabatic blocks further heat transfer between the bodies. Because the pair is also isolated from the surroundings, each temperature then remains fixed during the interval \(t_1\) to \(t_2\). Restoring the conducting wall at \(t_2\) restores the thermal path while the temperature difference is still present. Net heat flow therefore begins again in the same hotter-to-colder direction. The two curves resume their convergence until a common equilibrium temperature is eventually attained. No instantaneous equalisation occurs merely because a wall is changed. The correct graph must therefore show convergence, an unequal-temperature plateau interval, and renewed convergence.
