FOCUSED REVIEW
Focused Review — Thermal Expansion — Algebra Based
Review thermal expansion through physical meaning, essential relationships, representative calculations, and applications.
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-U02 | TOPIC: Thermal Expansion | COURSE LEVEL: Algebra-Based College Physics
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
Key Ideas
Linear Expansion Model
For a uniform isotropic material over a moderate temperature range, ΔL = αL0ΔT and L = L0(1 + αΔT). The units of α are inverse temperature, such as °C⁻¹ or K⁻¹.
ACTIVITY 2
Common Mistakes
A steel rail of length 20.0 m heated by 25°C with α = 12 × 10⁻⁶/°C expands 6.0 mm.
Keep the sign of ΔT through the calculation.
ACTIVITY 3
Quick Application
Compare two temperatures expressed on different scales.
Which is warmer: 50 °F or 15 °C? Convert before deciding.
Reveal Answers
50 °F = 10 °C, so 15 °C is warmer.
Why it works: A same-scale comparison prevents a misleading numerical comparison.
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
Linear Expansion Coefficients
Area and Volume Expansion
T(K) = T(°C) + 273.15; T(°F) = (9/5)T(°C) + 32
Example: −40 °C = −40 °F, but it equals 233.15 K.
Sensei note: Never add 273.15 to a temperature difference.
KEY CONCEPT 2
For small fractional changes, ΔA = 2αA0ΔT and ΔV = βV0ΔT. For isotropic solids, β ≈ 3α. Use initial area or volume in these linear approximations.
Thermal expansion requires equal temperature and no net heat transfer. The Thermal Expansion Model allows a expansion measurement to compare systems that never touch each other directly.
If A and C equilibrate, and B and C equilibrate, then temperature of A = temperature of B
Example: Two samples that settle at the same expansion measurement reading share a temperature even if their masses differ.
Sensei note: Equal temperature does not imply equal internal energy.
KEY CONCEPT 3
Linear Expansion Measurement Calibration
A thermometric property such as length or resistance can be calibrated between two known points. If its response is linear, equal fractions of the property range represent equal fractions of the temperature range.
T = T₁ + [(X − X₁)/(X₂ − X₁)](T₂ − T₁)
Example: A sensor halfway between its 0 °C and 100 °C calibration readings indicates 50 °C.
Sensei note: Linear interpolation is valid only when the response is stated or shown to be linear.
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
Guided Example
Carry out a two-step scale conversion.
A 0.100 m³ solid with α = 20 × 10⁻⁶/°C heated by 50°C gains about 3.00 × 10⁻⁴ m³.
PRACTICE 2
Independent Check
Use a linear expansion measurement calibration.
A liquid column is 2.0 cm at 0 °C and 12.0 cm at 100 °C. Assuming linear response, what temperature corresponds to 7.0 cm?
Reveal Answers
The fraction is (7−2)/(12−2)=0.50, so the temperature is 50 °C.
Why it works: A linear expansion measurement maps equal fractions of its range to equal temperature fractions.
PRACTICE 3
Independent Problem
Do not use α directly in the volume equation unless you first replace β with 3α.
A equilibrates with C at 290 K and B equilibrates with C at 290 K. Predict the initial transfer when A and B touch.
Reveal Answers
No initial net heat transfer occurs because A and B have equal temperatures.
Why it works: The Thermal Expansion Model transfers the equality through reference C.
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
Temperature Change
Convert an interval, not an absolute value.
Apparent Expansion and Thermal Stress
QUICK CHECK 2
A liquid and its container both expand. The observed overflow depends on the difference between their volume-expansion coefficients. If a solid is prevented from expanding, the thermal strain αΔT leads to thermal stress with magnitude Yα|ΔT| in the ideal fully constrained model.
Use the Thermal Expansion Model.
A and B each equilibrate with the same expansion measurement at 305 K. What happens initially if A touches B?
Reveal Answers
No initial net heat transfer; A and B have the same temperature.
Why it works: The shared reference temperature makes A and B mutually equilibrated.
QUICK CHECK 3
Calibrate a Expansion Measurement
Use linear interpolation.
A linear sensor reads X=4 at 10 °C and X=10 at 40 °C. What temperature corresponds to X=7?
Reveal Answers
X=7 is halfway from 4 to 10, so T is halfway from 10 °C to 40 °C: 25 °C.
Why it works: Both the sensor property and temperature are at the same 0.50 fraction.
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
Convert Values with Offset and Scale
A tightly constrained steel member can experience large stress even when its free expansion would be only millimeters.
KEY TAKEAWAY 2
Compare Temperatures Before Predicting Transfer
Heat flows spontaneously from higher to lower temperature until thermal expansion is reached.
KEY TAKEAWAY 3
Calibration Maps a Property to Temperature
A linear calibration uses the same fractional position on the property and temperature ranges.
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?
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