301. A heat engine absorbs \(2000\,\text{J}\) of heat from a hot reservoir and rejects \(1200\,\text{J}\) to a cold reservoir in one cycle. Its efficiency is
ⓐ. \(60\%\)
ⓑ. \(20\%\)
ⓒ. \(80\%\)
ⓓ. \(40\%\)
Correct Answer: \(40\%\)
Explanation: \( \textbf{Heat absorbed:} \) \(Q_1=2000\,\text{J}\).
\( \textbf{Heat rejected:} \) \(Q_2=1200\,\text{J}\).
\( \textbf{Work output per cycle:} \)
\[
W=Q_1-Q_2
\]
\[
W=2000-1200=800\,\text{J}
\]
\( \textbf{Efficiency definition:} \)
\[
\eta=\frac{W}{Q_1}
\]
\( \textbf{Substitution:} \)
\[
\eta=\frac{800}{2000}=0.40
\]
\( \textbf{Percentage form:} \)
\[
\eta=40\%
\]
\( \textbf{Final answer:} \) The heat engine has efficiency \(40\%\).
302. A heat engine has efficiency \(30\%\) and absorbs \(5000\,\text{J}\) of heat from the source in each cycle. The work delivered per cycle is
ⓐ. \(5000\,\text{J}\)
ⓑ. \(1500\,\text{J}\)
ⓒ. \(6500\,\text{J}\)
ⓓ. \(3500\,\text{J}\)
Correct Answer: \(1500\,\text{J}\)
Explanation: \( \textbf{Given efficiency:} \) \(\eta=30\%=0.30\).
\( \textbf{Heat absorbed:} \) \(Q_1=5000\,\text{J}\).
\( \textbf{Efficiency relation:} \)
\[
\eta=\frac{W}{Q_1}
\]
\( \textbf{Solve for work:} \)
\[
W=\eta Q_1
\]
\( \textbf{Substitution:} \)
\[
W=(0.30)(5000)\,\text{J}
\]
\( \textbf{Calculation:} \)
\[
W=1500\,\text{J}
\]
\( \textbf{Energy reading:} \) Only part of the absorbed heat appears as work; the rest is rejected to the cold reservoir.
\( \textbf{Final answer:} \) The work delivered per cycle is \(1500\,\text{J}\).
303. A cyclic engine absorbs \(Q_1\) from a source and rejects \(Q_2\) to a sink, where \(Q_1\) and \(Q_2\) are positive magnitudes. The work output is
ⓐ. \(Q_1+Q_2\)
ⓑ. \(Q_1-Q_2\)
ⓒ. \(Q_2-Q_1\)
ⓓ. \(\frac{Q_1}{Q_2}\)
Correct Answer: \(Q_1-Q_2\)
Explanation: In a complete cycle, the working substance returns to its initial state. Therefore, \(\Delta U_{\text{cycle}}=0\). The net heat absorbed by the engine is the heat taken from the hot source minus the heat rejected to the cold sink. Thus, the work output is \(W=Q_1-Q_2\). The expression \(Q_1+Q_2\) would incorrectly treat rejected heat as useful output. Heat rejected is still energy, but it is not converted into work by the engine.
304. A heat engine is claimed to absorb \(1000\,\text{J}\) of heat from a single hot reservoir and convert all of it into \(1000\,\text{J}\) of work in each cycle, with no other effect. This claim conflicts most directly with
ⓐ. Mayer relation for molar heat capacities
ⓑ. the zeroth-law statement of thermal equilibrium
ⓒ. the ideal-gas equation of state for \(PV=nRT\)
ⓓ. the Kelvin-Planck statement of the second law
Correct Answer: the Kelvin-Planck statement of the second law
Explanation: The Kelvin-Planck statement says that no heat engine operating in a cycle can convert all the heat absorbed from a single reservoir completely into work with no other effect. A real heat engine must reject some heat to a colder reservoir. The first law alone would allow energy balance \(Q=W\), but the second law restricts which cyclic conversions are possible. The issue is not temperature measurement or the gas equation. The claim describes \(100\%\) conversion of heat from one reservoir into work in a cycle, which is forbidden by the second law.
305. Study the table and choose the row that correctly describes a heat engine.
| Row | Heat from hot reservoir | Heat to cold reservoir | Work output |
|---|
| P | \(Q_1\) | \(Q_2\) | \(Q_1-Q_2\) |
| Q | \(Q_1\) | \(0\) | \(Q_1\) always possible in a cycle |
| R | \(Q_2\) | \(Q_1\) | \(Q_1+Q_2\) |
| S | \(0\) | \(Q_2\) | \(Q_2\) |
ⓐ. Row Q
ⓑ. Row S
ⓒ. Row P
ⓓ. Row R
Correct Answer: Row P
Explanation: A heat engine absorbs heat \(Q_1\) from a hot reservoir and rejects heat \(Q_2\) to a cold reservoir. Since the working substance completes a cycle, its net internal-energy change is zero. The work output is therefore \(W=Q_1-Q_2\). Row Q describes complete conversion of heat from one reservoir into work, which is not possible for a cyclic heat engine. Rows R and S do not represent the usual heat-engine energy flow.
306. A heat engine rejects \(600\,\text{J}\) of heat and produces \(400\,\text{J}\) of work in each cycle. How much heat does it absorb from the hot reservoir, and what is its efficiency?
ⓐ. \(Q_1=600\,\text{J}\), \(\eta=67\%\)
ⓑ. \(Q_1=1000\,\text{J}\), \(\eta=40\%\)
ⓒ. \(Q_1=200\,\text{J}\), \(\eta=200\%\)
ⓓ. \(Q_1=400\,\text{J}\), \(\eta=100\%\)
Correct Answer: \(Q_1=1000\,\text{J}\), \(\eta=40\%\)
Explanation: \( \textbf{Heat rejected:} \) \(Q_2=600\,\text{J}\).
\( \textbf{Work output:} \) \(W=400\,\text{J}\).
\( \textbf{Cycle energy relation:} \)
\[
W=Q_1-Q_2
\]
\( \textbf{Solve for heat absorbed:} \)
\[
Q_1=W+Q_2
\]
\[
Q_1=400+600=1000\,\text{J}
\]
\( \textbf{Efficiency:} \)
\[
\eta=\frac{W}{Q_1}
\]
\[
\eta=\frac{400}{1000}=0.40
\]
\( \textbf{Percentage form:} \)
\[
\eta=40\%
\]
\( \textbf{Final answer:} \) \(Q_1=1000\,\text{J}\) and \(\eta=40\%\).
307. A device works in a cycle between two reservoirs. It takes heat from a colder region, rejects heat to a hotter region, and requires external work input. This device is best described as
ⓐ. an isolated system
ⓑ. an isochoric gas process
ⓒ. a heat engine
ⓓ. a refrigerator
Correct Answer: a refrigerator
Explanation: A refrigerator transfers heat from a cold region to a hot region. Such transfer is opposite to the natural direction of heat flow, so external work input is required. The device operates in a cycle, but its purpose is not to deliver work output like a heat engine. Instead, work is supplied to remove heat from the cold space. The working substance returns to its initial state after a cycle, while heat is moved from cold to hot.
308. The coefficient of performance of a refrigerator is defined as
ⓐ. \(\beta=\frac{W}{Q_2}\)
ⓑ. \(\beta=\frac{Q_2}{W}\)
ⓒ. \(\beta=\frac{Q_1}{Q_2}\)
ⓓ. \(\beta=\frac{Q_1-Q_2}{Q_1}\)
Correct Answer: \(\beta=\frac{Q_2}{W}\)
Explanation: For a refrigerator, the useful effect is the heat removed from the cold region. This heat is usually denoted by \(Q_2\). The required input is the external work \(W\). Therefore, the coefficient of performance is \(\beta=\frac{Q_2}{W}\). It is not an efficiency because the useful output is not work; it is heat removed from the cold space.
309. A refrigerator removes \(800\,\text{J}\) of heat from its cold chamber in one cycle and requires \(200\,\text{J}\) of work input. Its coefficient of performance is
ⓐ. \(0.25\)
ⓑ. \(4\)
ⓒ. \(5\)
ⓓ. \(2\)
Correct Answer: \(4\)
Explanation: \( \textbf{Heat removed from cold chamber:} \) \(Q_2=800\,\text{J}\).
\( \textbf{Work input:} \) \(W=200\,\text{J}\).
\( \textbf{Coefficient of performance:} \)
\[
\beta=\frac{Q_2}{W}
\]
\( \textbf{Substitution:} \)
\[
\beta=\frac{800}{200}
\]
\( \textbf{Calculation:} \)
\[
\beta=4
\]
\( \textbf{Unit check:} \) Both \(Q_2\) and \(W\) are energies, so \(\beta\) has no unit.
\( \textbf{Final answer:} \) The coefficient of performance is \(4\).
310. A refrigerator extracts \(600\,\text{J}\) from a cold chamber and rejects \(750\,\text{J}\) to the surroundings in one cycle. The work input is
ⓐ. \(600\,\text{J}\)
ⓑ. \(1350\,\text{J}\)
ⓒ. \(750\,\text{J}\)
ⓓ. \(150\,\text{J}\)
Correct Answer: \(150\,\text{J}\)
Explanation: \( \textbf{Heat taken from cold chamber:} \) \(Q_2=600\,\text{J}\).
\( \textbf{Heat rejected to hot surroundings:} \) \(Q_1=750\,\text{J}\).
\( \textbf{Energy balance for refrigerator:} \)
\[
Q_1=Q_2+W
\]
\( \textbf{Solve for work input:} \)
\[
W=Q_1-Q_2
\]
\( \textbf{Substitution:} \)
\[
W=750-600
\]
\( \textbf{Calculation:} \)
\[
W=150\,\text{J}
\]
\( \textbf{Physical meaning:} \) The rejected heat is larger than the extracted heat because it includes the supplied work.
\( \textbf{Final answer:} \) The work input is \(150\,\text{J}\).
311. The Clausius statement of the second law says that
ⓐ. cold-to-hot flow occurs spontaneously in contact
ⓑ. internal energy is not conserved in thermal processes
ⓒ. all heat engines can have \(100\%\) efficiency
ⓓ. cold-to-hot heat flow needs external work
Correct Answer: cold-to-hot heat flow needs external work
Explanation: The Clausius statement describes the direction of heat transfer. It says that heat cannot, by itself, flow from a colder body to a hotter body without an external agency. Refrigerators do transfer heat from cold to hot, but they need work input. This does not violate the statement because there is an additional effect: work is supplied. The statement does not deny energy conservation; it adds a direction condition to thermal processes.
312. Assertion: A refrigerator does not violate the Clausius statement of the second law.
Reason: It transfers heat from a cold space to a warmer region only when external work is supplied.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The Assertion is true because refrigerators are allowed by the second law when work is supplied. The Clausius statement forbids heat transfer from cold to hot as the only effect. A refrigerator uses external work input to drive heat from the cold chamber to the warmer surroundings. The Reason gives exactly the extra effect that makes the process possible. Without work input, the same heat transfer direction would not occur naturally.
313. A heat pump supplies heat \(Q_1\) to a warm room by taking heat \(Q_2\) from a cold outside region and using work \(W\). Its coefficient of performance as a heat pump is
ⓐ. \(\frac{Q_2}{W}\)
ⓑ. \(\frac{Q_1-Q_2}{Q_2}\)
ⓒ. \(\frac{Q_1}{W}\)
ⓓ. \(\frac{W}{Q_1}\)
Correct Answer: \(\frac{Q_1}{W}\)
Explanation: For a heat pump, the useful effect is the heat delivered to the warm region. This quantity is \(Q_1\). The input needed to operate the pump is work \(W\). Therefore, the heat-pump coefficient of performance is \(\frac{Q_1}{W}\). This differs from a refrigerator, where the useful effect is heat removed from the cold region, \(\frac{Q_2}{W}\).
314. A heat pump delivers \(3000\,\text{J}\) of heat to a room while consuming \(750\,\text{J}\) of work. Its coefficient of performance is
ⓐ. \(0.25\)
ⓑ. \(2.25\)
ⓒ. \(5.0\)
ⓓ. \(4.0\)
Correct Answer: \(4.0\)
Explanation: \( \textbf{Heat delivered to room:} \) \(Q_1=3000\,\text{J}\).
\( \textbf{Work input:} \) \(W=750\,\text{J}\).
\( \textbf{Heat-pump coefficient of performance:} \)
\[
\text{COP}_{\text{HP}}=\frac{Q_1}{W}
\]
\( \textbf{Substitution:} \)
\[
\text{COP}_{\text{HP}}=\frac{3000}{750}
\]
\( \textbf{Calculation:} \)
\[
\text{COP}_{\text{HP}}=4.0
\]
\( \textbf{Interpretation:} \) The delivered heat can exceed the work input because some heat is taken from the colder outside region.
\( \textbf{Final answer:} \) The coefficient of performance is \(4.0\).
315. A refrigerator has \(\beta=5\) and removes \(1000\,\text{J}\) of heat from the cold chamber in each cycle. The work input required per cycle is
ⓐ. \(100\,\text{J}\)
ⓑ. \(6000\,\text{J}\)
ⓒ. \(5000\,\text{J}\)
ⓓ. \(200\,\text{J}\)
Correct Answer: \(200\,\text{J}\)
Explanation: \( \textbf{Coefficient of performance:} \) \(\beta=5\).
\( \textbf{Heat removed from cold chamber:} \) \(Q_2=1000\,\text{J}\).
\( \textbf{Refrigerator COP relation:} \)
\[
\beta=\frac{Q_2}{W}
\]
\( \textbf{Solve for work:} \)
\[
W=\frac{Q_2}{\beta}
\]
\( \textbf{Substitution:} \)
\[
W=\frac{1000}{5}
\]
\( \textbf{Calculation:} \)
\[
W=200\,\text{J}
\]
\( \textbf{Final answer:} \) The work input required per cycle is \(200\,\text{J}\).
316. The main difference between heat-engine efficiency and refrigerator coefficient of performance is that
ⓐ. efficiency has unit \(\text{J}\), while COP has unit \(\text{K}\)
ⓑ. both are always less than \(1\)
ⓒ. efficiency uses work output; COP uses heat removed
ⓓ. COP applies only when no work is supplied
Correct Answer: efficiency uses work output; COP uses heat removed
Explanation: A heat engine is designed to produce work, so its efficiency is \(\eta=\frac{W}{Q_1}\). A refrigerator is designed to remove heat from a cold region, so its coefficient of performance is \(\beta=\frac{Q_2}{W}\). Both quantities are ratios of energies and are therefore dimensionless. A refrigerator COP can be greater than \(1\), unlike heat-engine efficiency, which is less than \(1\) for any real cyclic engine. The two ratios measure different useful effects.
317. A cyclic device takes \(900\,\text{J}\) from a cold body and rejects \(1200\,\text{J}\) to a hot body. The most suitable interpretation is
ⓐ. it is an isolated system with no energy exchange
ⓑ. it violates the first law because \(1200\,\text{J}\) is greater than \(900\,\text{J}\)
ⓒ. it is a refrigerator or heat pump requiring \(300\,\text{J}\) of work input
ⓓ. it is a heat engine producing \(300\,\text{J}\) of work
Correct Answer: it is a refrigerator or heat pump requiring \(300\,\text{J}\) of work input
Explanation: The device transfers heat from a cold body to a hot body. This direction requires work input, so the device behaves as a refrigerator or heat pump depending on the useful effect. The heat rejected to the hot body is \(1200\,\text{J}\), while the heat extracted from the cold body is \(900\,\text{J}\). The difference must be supplied as work: \(W=1200-900=300\,\text{J}\). This is consistent with the first law because the rejected heat includes both extracted heat and work input.
318. Consider the following statements.
I. A heat engine operating in a cycle has \(\Delta U_{\text{cycle}}=0\).
II. A refrigerator requires work input to transfer heat from cold to hot.
III. A heat engine can convert all absorbed heat into work if friction is absent.
ⓐ. I, II, and III
ⓑ. II and III only
ⓒ. I and II only
ⓓ. I and III only
Correct Answer: I and II only
Explanation: Statement I is true because the working substance returns to its initial state after each cycle, so its internal-energy change over a cycle is zero. Statement II is true because transferring heat from cold to hot needs external work input. Statement III is false because absence of friction alone does not allow complete conversion of heat into work in a cyclic heat engine. The second law requires rejection of some heat to a cold reservoir. A reversible engine can be ideal, but it still cannot have \(100\%\) efficiency between finite reservoirs.
319. A heat engine absorbs \(Q_1=2500\,\text{J}\) and has efficiency \(0.32\). The heat rejected to the cold reservoir is
ⓐ. \(1700\,\text{J}\)
ⓑ. \(2500\,\text{J}\)
ⓒ. \(800\,\text{J}\)
ⓓ. \(3300\,\text{J}\)
Correct Answer: \(1700\,\text{J}\)
Explanation: \( \textbf{Heat absorbed:} \) \(Q_1=2500\,\text{J}\).
\( \textbf{Efficiency:} \) \(\eta=0.32\).
\( \textbf{Work output:} \)
\[
W=\eta Q_1
\]
\[
W=(0.32)(2500)=800\,\text{J}
\]
\( \textbf{Engine energy relation:} \)
\[
W=Q_1-Q_2
\]
\( \textbf{Solve for rejected heat:} \)
\[
Q_2=Q_1-W
\]
\[
Q_2=2500-800=1700\,\text{J}
\]
\( \textbf{Physical reading:} \) The rejected heat is the part of absorbed heat not converted into work.
\( \textbf{Final answer:} \) \(Q_2=1700\,\text{J}\).
320. A proposed refrigerator removes \(500\,\text{J}\) from a cold chamber and rejects \(500\,\text{J}\) to a hotter room with no work input. The proposal is impossible mainly because
ⓐ. heat cannot be measured in \(\text{J}\)
ⓑ. refrigerators cannot operate in cycles
ⓒ. it would conserve energy too exactly
ⓓ. cold-to-hot transfer without work
Correct Answer: cold-to-hot transfer without work
Explanation: The proposed device removes heat from a cold chamber and delivers the same amount to a hotter room without work input. This would make heat pass from cold to hot as the only effect. The Clausius statement of the second law rules out such a process. A real refrigerator needs work input, and then the heat rejected to the room is greater than the heat removed from the cold chamber. The problem is not energy conservation alone; it is the forbidden direction of heat transfer without compensation.
