FE Mechanical Practice Question Solution included after investigation
CASE E-053 Β· FE

The Refrigerator With a COP Greater Than One

The refrigerator removed 6 kW of heat using 2 kW of work. A COP of 3 was correct, not impossible.

β–Ά INTERACTIVE CASE FILE β€’ 5–10 MIN β€’ CLICK TO INVESTIGATE

Inspect, commit, prove, fix, and sign off.

Case E-053 engineering failure visual
HOW THIS WORKS

Think like
a real engineer.

Follow the investigation process used in the field β€” in five guided steps.

1

Inspect

Review the problem, diagram, and evidence.

2

Commit

Choose your hypothesis.

3

Prove

Run calculations and test your idea.

4

Fix

Select and validate a safe correction.

5

Sign Off

See the full debrief and key takeaways.

Case Brief

SAME PRINCIPLES. HIGHER STANDARDS.
PROBLEM STATEMENT

A refrigerator removes 6 kW of heat from a cold space while consuming 2 kW of electrical work. A reviewer rejects the calculated COP of 3 because it exceeds 1. Determine the correct COP and condenser heat rejection, then explain why the result does not violate the first law.

Thermodynamics Mechanical Design Verification Root Cause Analysis Safe Correction
E-053
INVESTIGATION WORKSPACE WORK THROUGH THE CASE β€” YOUR CHOICES MATTER.
πŸ”’
DEBRIEF (LOCKED)

Complete the investigation
to unlock the full debrief.

Find the root cause, confirm the fix, and see how this connects to the exam.

πŸ”’Root causeComplete the case
to reveal
πŸ”’Correct fixComplete the case
to reveal
πŸ”’Exam takeawayComplete the case
to reveal
CASE CLOSED Β· ENGINEERING VERDICT

What actually failed β€” and what should change.

What failed

COP was interpreted as if it were a thermal efficiency.

What was missed

The refrigerator moves heat in addition to consuming work.

Corrected model

Use COP_R=QL/W and QH=QL+W.

QUICK ANSWER

Refrigerator COP is QL/W. Removing 6 kW with 2 kW of work gives COP_R=3. The condenser rejects QH=QL+W=8 kW, so energy is conserved even though COP exceeds 1.

ENGINEERING TAKEAWAY

Refrigerator COP measures heat moved per unit work input, not the fraction of input energy converted to useful output. COP_R can exceed 1.

CASE QUESTIONS

Questions this case should settle.

Can refrigerator COP be greater than 1?

Yes. COP measures heat moved per unit work input, not conversion efficiency.

If QL=6 kW and W=2 kW, what heat is rejected?

QH=8 kW.

NEXT FE PRACTICE QUESTION

The Turbine That Produced 1.2 MW From an Enthalpy Drop

Thermodynamics Β· 5–10 min

Continue Practice β†’
WHAT YOU’LL LEARN

Real failures. Lasting skills.

Each case is designed to build the judgment, analysis, and confidence you need for engineering exams β€” and beyond.

  • Apply core engineering concepts to real-world problems
  • Practice structured troubleshooting and analysis
  • Strengthen exam-ready thinking through realistic scenarios
COMMON QUESTIONS
Do I need prior knowledge to use these cases?

Basic subject familiarity helps, but every case is designed to teach through the investigation itself.

How long does a case take?

Most cases are designed for a focused 5–10 minute investigation.

Are the cases aligned with engineering exams?

Each case is mapped to a verified exam, subject, topic, and misconception before publication.

What happens after I complete a case?

The sealed debrief unlocks with the root cause, corrected reasoning, fix, and takeaway.