FULL REVIEW
Full Review — Heat Engines, Refrigerators, and Heat Pumps — Foundational
Review the essential ideas, relationships, and problem-solving tools for Heat Engines, Refrigerators, and Heat Pumps.
TIME
45–60 minutes
BEST FOR
A complete unit review
FINISH WITH
A readiness check
After this full review, you'll be able to...
recall the essential ideas, apply them to representative problems, and determine what to study next.
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: Foundational
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 45–60 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 the energy transfers in a heat engine.
Draw hot and cold reservoirs. Label incoming QH, outgoing QC, and work Wout.
Reveal Answers
Three arrows: QH in, Wout out, QC out.
Why it works: The working fluid returns to its original state after a cycle.
ACTIVITY 2
Warm-up 2: Useful output
Classify useful outputs.
Explain which transfer counts as useful for an engine, refrigerator, and heat pump.
Reveal Answers
Engine: work; refrigerator: cold-space heat removal; heat pump: warm-space heat delivery.
Why it works: Useful output depends on the device's job.
ACTIVITY 3
Warm-up 3: Cycle balance
Apply conservation to a cycle.
An engine absorbs 800 J and rejects 600 J. Find work 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: First-law cycle balance gives work as the difference.
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
Let's rebuild the key ideas one step at a time. Focus on understanding the relationships before worrying about solving problems.
KEY CONCEPT 1
Heat engines and the first law
A heat engine takes energy from a hot source, turns part of it into useful work, and releases the rest to a cooler place. Energy is conserved: heat in = work out + heat rejected.
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: For a full cycle, internal energy returns to its initial value; never equate heat input to work alone.
KEY CONCEPT 2
Refrigerators and heat pumps
A refrigerator uses work to move heat from a cold space to a warm room. A heat pump uses the same cycle, but its useful output is the heat delivered to the warm space.
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: The two COP definitions differ because cooling and heating select different useful outputs.
KEY CONCEPT 3
Performance and limits
Engine efficiency = useful work ÷ heat taken from the hot source. Refrigerator COP = heat removed from the cold space ÷ work input. Heat-pump COP = heat delivered to the warm space ÷ work input.
Reversible bound: ηCarnot = 1 − TC/TH, with kelvin temperatures.
Example: At 500 K and 300 K, the reversible engine limit is 1 − 300/500 = 0.40. A claim of 50% efficiency between these reservoirs violates that limit.
Sensei note: Use kelvin in reservoir-temperature ratios; ideal bounds do not describe every real machine.
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
Now it's time to apply what you've reviewed.
Work through each activity in order. The examples become gradually more challenging, and each one prepares you for the final readiness check.
PRACTICE 1
Practice 1: Draw flows, subtract rejected heat, then compute efficiency.
Draw flows, subtract rejected heat, then compute efficiency.
An engine absorbs 1,000 J and rejects 700 J. Wout = 300 J and η = 300/1,000 = 0.30.
Reveal Answers
Wout = 300 J; η = 30%.
Why it works: Compute 1,000 − 700, then divide work by 1,000 J.
PRACTICE 2
Practice 2: Use separate cooling and heating numerators.
Use separate cooling and heating numerators.
A refrigerator removes 450 J with 150 J work. Find delivered heat, cooling COP, and heating COP. Identify the useful transfer for each ratio.
Reveal Answers
QH = 600 J; COPR = 3; COPHP = 4.
Why it works: Add 450 and 150; use each useful heat transfer over 150 J.
PRACTICE 3
Practice 3: Compare an efficiency claim with a reversible limit.
Compare an efficiency claim with a reversible limit.
An engine operates between 500 K and 300 K. Could it deliver 500 J of work from 1,000 J of hot-source heat? Compare its claimed efficiency with the reversible limit.
Reveal Answers
No: 50% exceeds the 40% reversible limit.
Why it works: A real engine cannot exceed the reversible bound between the same reservoirs.
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
Energy balance
Calculate work before performance.
An engine absorbs 900 J and rejects 630 J. Find Wout and η.
Reveal Answers
Wout = 270 J; η = 270/900 = 0.30.
Why it works: Subtract 630 J from 900 J and divide by 900 J.
QUICK CHECK 2
Reversed cycle
Use separate cooling and heating numerators.
A device draws 600 J from the cold reservoir using 200 J of work. Find QH, COPR, and COPHP.
Reveal Answers
QH = 800 J; COPR = 600/200 = 3; COPHP = 800/200 = 4.
Why it works: The delivered heat includes cold-side transfer plus work.
QUICK CHECK 3
Temperature limit
Use absolute temperatures.
For reservoirs at 600 K and 300 K, find the maximum reversible engine efficiency. Is a 60% claim possible?
Reveal Answers
ηCarnot = 1 − 300/600 = 0.50; a 60% claim is impossible.
Why it works: Use absolute reservoir temperatures and compare with the reversible upper bound.
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 balance
QH = Wout + QC; η = Wout/QH.
KEY TAKEAWAY 2
Cooling and heating
QH = QC + Win; COPR = QC/Win and COPHP = QH/Win.
KEY TAKEAWAY 3
Physical limits
The reversible temperature bound is an upper limit; real cyclic devices perform below it.
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Next Step
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