FOCUSED REVIEW
Focused Review — Heat Engines, Refrigerators, and Heat Pumps — Algebra-Based
Reinforce the highest-leverage ideas and representative problem-solving tools for Heat Engines, Refrigerators, and Heat Pumps.
TIME
Approximately 15 minutes
BEST FOR
Targeted reinforcement
FINISH WITH
A readiness check
After this focused review, you'll be able to...
reinforce the key relationships, apply them to representative problems, and identify what still needs work.
Choose how you want to review
Course Alignment
This Physics Sensei Unit Review is an independent learning resource. Use it to reinforce key concepts, prepare for homework, or review before a quiz or exam.
RESOURCE: Physics Sensei Unit Review | UNIT ID: THM-U10 | TOPIC: Heat Engines, Refrigerators, and Heat Pumps | COURSE LEVEL: Algebra-Based
BEST USED ✓ After learning the unit ✓ Before starting homework ✓ Before a quiz or exam
Your Review Plan
Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.
6 Stages • Approximately 15 minutes.
Warm-Up Check
Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.
ACTIVITY 1
Warm-up 1: Energy-flow sketch
Identify hot-source heat, rejected heat, and work for an engine.
Draw three arrows and label the 800 J heat input, 600 J heat rejection, and unknown work output.
Reveal Answers
Work output is 200 J.
Why it works: Subtract rejected heat from absorbed heat: 800 − 600 = 200 J.
ACTIVITY 2
Warm-up 2: Useful output
Distinguish the useful output of a refrigerator from that of a heat pump.
In the same reversed cycle, which heat flow is useful for cooling, and which for heating?
Reveal Answers
Refrigeration uses QC; heating uses QH.
Why it works: The same reversed cycle can be described by which side is useful.
ACTIVITY 3
Warm-up 3: Cycle balance
Apply the energy balance to a short cycle.
An engine absorbs 800 J and rejects 600 J. Find its work output and efficiency.
Reveal Answers
A device receives 800 J from a hot source and rejects 600 J. Its work output is 200 J and efficiency is 200/800 = 0.25, or 25%.
Why it works: The useful work is 200 J, then 200/800 = 0.25.
Ready to strengthen your understanding?
You've refreshed what you already know. Next, you'll reinforce the essential concepts that will help you solve problems with confidence. Need to see the learning path again?
Core Concepts
Reinforce the two highest-leverage relationships, then use them in representative situations.
KEY CONCEPT 1
Heat engine energy balance
Using positive heat-transfer magnitudes, a cyclic heat engine satisfies QH = Wout + QC and η = Wout/QH = 1 − QC/QH. The heat rejected to the cold reservoir is part of the energy balance.
Engine: QH = Wout + QC; η = Wout/QH.
Example: A device receives 800 J from a hot source and rejects 600 J. Its work output is 200 J and efficiency is 200/800 = 0.25, or 25%.
Sensei note: Rejected heat cannot be omitted; it accounts for energy that does not become work.
KEY CONCEPT 2
Refrigerators and heat pumps
A refrigerator or heat pump requires work input: QH = QC + Win. COPR = QC/Win measures cooling; COPHP = QH/Win measures heating. Consequently COPHP = COPR + 1 for the same device. For reservoirs at TH and TC in kelvin, an ideal reversible limit is ηCarnot = 1 − TC/TH. An actual engine has lower efficiency. A coefficient of performance may exceed 1 because heat is moved as well as supplied by work.
Reversed cycle: QH = QC + Win; COPR = QC/Win; COPHP = QH/Win.
Example: A refrigerator removes 300 J from a cold compartment using 100 J of work. It delivers 400 J to the room; COPR = 300/100 = 3 and COPHP = 400/100 = 4.
Sensei note: Cooling COP uses removed heat; heating COP uses delivered heat.
Ready to apply these ideas?
You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?
Guided Practice
Apply the reinforced ideas to two representative situations, then use the strategy card to check your setup.
PRACTICE 1
Practice 1: Identify the useful transfer and calculate its ratio to the input.
Identify the useful transfer and calculate its ratio to the input.
For 800 J absorbed and 600 J rejected, Wout = 200 J; η = 0.25. Sketch the heat and work arrows before dividing.
Reveal Answers
A device receives 800 J from a hot source and rejects 600 J. Its work output is 200 J and efficiency is 200/800 = 0.25, or 25%.
Why it works: Calculate the work first, then divide by incoming heat.
PRACTICE 2
Practice 2: Solve a reversed-cycle energy balance.
Solve a reversed-cycle energy balance.
A heat pump delivers 1,200 J indoors using 300 J of work. Find outdoor heat drawn and the heating COP.
Reveal Answers
A heat pump delivers 1,200 J indoors while using 300 J of work. It draws 900 J from outdoors and has heating COP 1,200/300 = 4.
Why it works: Outdoor heat plus work gives delivered indoor heat.
Ready to check your understanding?
You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?
Confidence Check
You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.
QUICK CHECK 1
Engine efficiency
Show the energy balance and ratio.
An engine receives 1,000 J and rejects 700 J. Find its work output and efficiency.
Reveal Answers
Wout = 300 J; η = 300/1,000 = 0.30, or 30%.
Why it works: Conservation gives 1,000 − 700 = 300 J before the efficiency ratio.
QUICK CHECK 2
Cooling versus heating
Name the useful heat transfer in each mode.
A refrigerator removes 450 J using 150 J of work. Find the heat delivered to the room, COPR, and COPHP.
Reveal Answers
QH = 600 J; COPR = 450/150 = 3; COPHP = 600/150 = 4.
Why it works: The room receives the extracted heat plus work input.
How did it go?
You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?
Summary
Before moving on, take one final look at the most important ideas from this review.
KEY TAKEAWAY 1
Engine
Work output is the difference between absorbed and rejected heat; efficiency compares work to absorbed heat.
KEY TAKEAWAY 2
Reversed cycle
Refrigerator and heat-pump COP use different useful heat transfers for the same work input.
Ready for your next step?
You've reviewed the essential ideas one last time. Now choose the resource that best matches how confident you feel. Need a quick reminder?
Next Step
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