Thermal Properties

Materials Science · FE Reference Handbook section

Materials Science

4 formulas

10 exam-style examples

~53 min

All Materials Science lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Thermal Properties within Materials Science. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what thermal properties describes physically and when it applies.
State every one of the 4 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: percent versus fraction in composition and strain.

Lecture

Why this section exists. Thermal Properties is the part of Materials Science that lets you connect a steel, concrete or polymer specimen under test to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a definition, a phase-diagram read, or a one-line property calculation. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. percent versus fraction in composition and strain. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Concrete cylinder under axial load in a compression testing machine. — Photo 1. Where this shows up in practice: thermal properties.
Wikimedia Commons, public domain

Materials Science — Thermal Properties: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a steel, concrete or polymer specimen under test. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 4 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

α	Quantity produced by "α = thermal expansion coefficient" — read its definition and unit from the handbook line directly above the equation.
ε	Quantity produced by "ε = engineering strain" — read its definition and unit from the handbook line directly above the equation.
∆T	Quantity produced by "∆T = change in temperature" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The thermal expansion coefficient is the ratio of engineering strain to the change in temperature.
where
Specific heat (also called heat capacity) is the amount of heat required to raise the temperature of something or an amount of
something by 1 degree.
At constant pressure the amount of heat (Q) required to increase the temperature of something by ∆T is Cp∆T, where Cp is the
constant pressure heat capacity.
At constant volume the amount of heat (Q) required to increase the temperature of something by ∆T is Cv∆T, where Cv is the
constant volume heat capacity.
An object can have a heat capacity that would be expressed as energy/degree.
The heat capacity of a material can be reported as energy/degree per unit mass or per unit volume.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Steady heat flow through a wall — Thermal Properties

A 1.0 ft thick concrete wall of area 390 ft² has k = 1.7 Btu/(hr·ft·°F) and a 65°F temperature difference. What is the heat flow rate?

Given

k = 1.7 Btu/hr·ft·°F
A = 390 ft²
ΔT = 65°F
t = 1.0 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 1.7 (390) (65) /1.0$
Evaluate
$q = 43, 095 B t u / h r$

Answer: q ≈ 43,095 Btu/hr

Why the other options are there

43,095 Btu/hr (thickness multiplied)
110.5 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Thermal Properties

Example 2

Steady heat flow through a wall — Thermal Properties (2)

A 1.1 ft thick concrete wall of area 188 ft² has k = 0.8 Btu/(hr·ft·°F) and a 47°F temperature difference. What is the heat flow rate?

Given

k = 0.8 Btu/hr·ft·°F
A = 188 ft²
ΔT = 47°F
t = 1.1 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 0.8 (188) (47) /1.1$
Evaluate
$q = 6, 426 B t u / h r$

Answer: q ≈ 6,426 Btu/hr

Why the other options are there

7,776 Btu/hr (thickness multiplied)
37.6 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Thermal Properties

Example 3

Steady heat flow through a wall — Thermal Properties (3)

A 0.9 ft thick concrete wall of area 183 ft² has k = 1.6 Btu/(hr·ft·°F) and a 82°F temperature difference. What is the heat flow rate?

Given

k = 1.6 Btu/hr·ft·°F
A = 183 ft²
ΔT = 82°F
t = 0.9 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 1.6 (183) (82) /0.9$
Evaluate
$q = 26, 677 B t u / h r$

Answer: q ≈ 26,677 Btu/hr

Why the other options are there

21,609 Btu/hr (thickness multiplied)
131.2 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Thermal Properties

Example 4

Steady heat flow through a wall — Thermal Properties (4)

A 0.7 ft thick concrete wall of area 199 ft² has k = 1.8 Btu/(hr·ft·°F) and a 60°F temperature difference. What is the heat flow rate?

Given

k = 1.8 Btu/hr·ft·°F
A = 199 ft²
ΔT = 60°F
t = 0.7 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 1.8 (199) (60) /0.7$
Evaluate
$q = 30, 703 B t u / h r$

Answer: q ≈ 30,703 Btu/hr

Why the other options are there

15,044 Btu/hr (thickness multiplied)
108.0 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Thermal Properties

Example 5

Steady heat flow through a wall — Thermal Properties (5)

A 1.2 ft thick concrete wall of area 297 ft² has k = 1.2 Btu/(hr·ft·°F) and a 47°F temperature difference. What is the heat flow rate?

Given

k = 1.2 Btu/hr·ft·°F
A = 297 ft²
ΔT = 47°F
t = 1.2 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 1.2 (297) (47) /1.2$
Evaluate
$q = 13, 959 B t u / h r$

Answer: q ≈ 13,959 Btu/hr

Why the other options are there

20,101 Btu/hr (thickness multiplied)
56.4 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Thermal Properties

Example 6

Steady heat flow through a wall — Thermal Properties (6)

A 1.3 ft thick concrete wall of area 218 ft² has k = 1.0 Btu/(hr·ft·°F) and a 21°F temperature difference. What is the heat flow rate?

Given

k = 1.0 Btu/hr·ft·°F
A = 218 ft²
ΔT = 21°F
t = 1.3 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 1.0 (218) (21) /1.3$
Evaluate
$q = 3, 522 B t u / h r$

Answer: q ≈ 3,522 Btu/hr

Why the other options are there

5,951 Btu/hr (thickness multiplied)
21.0 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Thermal Properties

Example 7

Steady heat flow through a wall — Thermal Properties (7)

A 0.8 ft thick concrete wall of area 230 ft² has k = 0.8 Btu/(hr·ft·°F) and a 47°F temperature difference. What is the heat flow rate?

Given

k = 0.8 Btu/hr·ft·°F
A = 230 ft²
ΔT = 47°F
t = 0.8 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 0.8 (230) (47) /0.8$
Evaluate
$q = 10, 810 B t u / h r$

Answer: q ≈ 10,810 Btu/hr

Why the other options are there

6,918 Btu/hr (thickness multiplied)
37.6 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Thermal Properties

Example 8

Steady heat flow through a wall — Thermal Properties (8)

A 1.2 ft thick concrete wall of area 214 ft² has k = 0.7 Btu/(hr·ft·°F) and a 56°F temperature difference. What is the heat flow rate?

Given

k = 0.7 Btu/hr·ft·°F
A = 214 ft²
ΔT = 56°F
t = 1.2 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 0.7 (214) (56) /1.2$
Evaluate
$q = 6, 991 B t u / h r$

Answer: q ≈ 6,991 Btu/hr

Why the other options are there

10,067 Btu/hr (thickness multiplied)
39.2 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Thermal Properties

Example 9

Steady heat flow through a wall — Thermal Properties (9)

A 1.0 ft thick concrete wall of area 260 ft² has k = 2.2 Btu/(hr·ft·°F) and a 66°F temperature difference. What is the heat flow rate?

Given

k = 2.2 Btu/hr·ft·°F
A = 260 ft²
ΔT = 66°F
t = 1.0 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 2.2 (260) (66) /1.0$
Evaluate
$q = 37, 752 B t u / h r$

Answer: q ≈ 37,752 Btu/hr

Why the other options are there

37,752 Btu/hr (thickness multiplied)
145.2 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Thermal Properties

Example 10

Steady heat flow through a wall — Thermal Properties (10)

A 1.1 ft thick concrete wall of area 191 ft² has k = 1.2 Btu/(hr·ft·°F) and a 82°F temperature difference. What is the heat flow rate?

Given

k = 1.2 Btu/hr·ft·°F
A = 191 ft²
ΔT = 82°F
t = 1.1 ft

Find

Heat flow q

Start with the thinking

Fourier's law for a plane wall is linear in ΔT.
Thickness divides, area multiplies.

Step-by-step solution

Fourier — q = kAΔT/t
Substituting
$q = 1.2 (191) (82) /1.1$
Evaluate
$q = 17, 086 B t u / h r$

Answer: q ≈ 17,086 Btu/hr

Why the other options are there

20,674 Btu/hr (thickness multiplied)
98.4 Btu/hr (area omitted)

Reference: FE Reference Handbook — Materials Science → Thermal Properties

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a steel, concrete or polymer specimen under test, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Thermal Properties contains 4 relations; you must be able to find this page in under 15 seconds.
Exam style: a definition, a phase-diagram read, or a one-line property calculation.
Unit rule: percent versus fraction in composition and strain.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

percent versus fraction in composition and strain
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.