Why Polymers Soften Without Breaking Covalent Bonds

A plastic strip becomes soft when heated. You bend it into a new shape, allow it to cool, and it becomes rigid again. But if the polymer contains strong covalent bonds, how did it change shape without those bonds breaking?

The answer is that softening usually involves movement of polymer chains, not the breaking of their covalent backbones. Heating allows sections of the chains to change position and, in suitable polymers, allows neighbouring chains to slide past one another. The bonds connecting atoms within each chain can remain intact throughout.

There is a useful question hidden in this explanation: how could you actually tell whether a heated polymer had become more mobile or had started to break down chemically? Watching it soften is not enough.

01What changes when a polymer softens?

Consider a long polyethene molecule.

Its backbone consists of carbon atoms connected by covalent bonds. These bonds hold the atoms together as one continuous molecular chain.

Now imagine several of these chains packed closely together.

There are two different types of interaction to consider:

• Within each chain: strong covalent bonds connect the atoms.
• Between neighbouring chains: intermolecular attractions influence how closely the chains pack and how easily they move relative to one another.

In non-polar polymers such as polyethene, London dispersion forces are an important attraction between chains. Polar polymers can also have dipole-dipole attractions or hydrogen bonding, depending on their chemical structures.

If you need to refresh that distinction, see Covalent Molecular Structures and Intermolecular Forces.

Predict what heating will do

Suppose the temperature increases, but remains below the temperature at which significant chemical degradation occurs.

Which is more likely?

1. The carbon-carbon bonds along the polymer backbone begin breaking, producing shorter molecules.
2. Sections of the existing chains become more mobile, changing their arrangements without breaking those bonds.

For ordinary thermal softening, the second explanation is the appropriate starting model.

A polymer chain is not a completely rigid rod. Rotation around suitable single bonds allows sections of the chain to adopt different three-dimensional arrangements. These arrangements are called conformations.

Heating increases the ability of chain segments to undergo these movements. Neighbouring chains can also rearrange relative to one another when the surrounding structure permits it.

Imagine several long shoelaces tangled together. You can move and rearrange the laces without cutting them. However, their tangles make some movements difficult.

The shoelaces represent polymer chains, while the restrictions created by their surroundings represent, very roughly, the effects of intermolecular attractions and entanglements.

The analogy has limits. Polymer chains are molecular structures, not ropes. Their movement depends on bond rotation, thermal motion, intermolecular interactions, and the surrounding arrangement of atoms. Heating doesn’t simply remove all the attractions between them.

Why increased movement changes stiffness

A rigid plastic resists bending because its molecular structure restricts rearrangement.

When chain segments become more mobile, the material may deform further under the same applied force.

This is a change in a bulk property: something measurable about the whole material, such as stiffness, flexibility, or resistance to deformation.

It does not automatically mean the chemical identity of its polymer molecules has changed.

There is another complication. Polymers can be viscoelastic, meaning their deformation depends on both the applied force and the time for which it acts. A material might bend slightly during the first second of loading, then continue to bend over the following minute.

That gradual deformation is called creep. It becomes particularly important in many polymers at higher temperatures.

So when someone says a polymer has become softer, we need to ask what was measured, at what temperature, under what force, and for how long.

02An experiment: measure softening without assuming bond breaking

Imagine investigating a thermoplastic strip that becomes noticeably easier to bend when warmed.

The first question is straightforward:

Does a polymer become easier to deform as its temperature increases, even when the applied force stays constant?

The apparatus

Use a controlled laboratory heating chamber containing a single polymer strip.

The strip is clamped horizontally at its left end, leaving an 80 mm length projecting towards the right.

A small weight is attached to the free end. Its downward force is \(0.20\ \mathrm{N}\).

A temperature probe measures the temperature near the strip, while a displacement scale measures how far its free end moves downwards.

The arrangement is:

• The left clamp holds one end of the strip fixed.
• The free end points horizontally towards the right before loading.
• The weight pulls the free end downwards.
• The displacement sensor measures the resulting downward deflection.
• The temperature is changed while the strip’s dimensions, loading arrangement, and measurement time remain controlled.

At each temperature, measure the tip position without the weight, then apply the weight for 30 seconds and record the additional downward movement.

Call this movement \(\delta\), measured in millimetres (mm). Take downward deflection as positive.

Using the unloaded position as the reference is important. A heated strip might change its resting shape even without a weight attached.

What should the original experiment show?

If heating increases chain mobility, the strip should generally bend further under the same force.

If substantial backbone breaking were responsible for the change, we would expect evidence that the polymer molecules had become chemically altered, although bending measurements alone would not reveal that.

Consider these hypothetical results.

Illustrative data only – not measurements from a performed experiment. Assume the strip remains chemically stable throughout the trial.

Every deflection is measured relative to the unloaded position at that temperature.

The important observation is the large increase in deformation between 25°C and 70°C.

After cooling, the strip’s response becomes much closer to its original response.

A simple explanation is that heating increased molecular mobility, making deformation easier. Cooling restricted that movement again.

However, the final value is 2.6 mm rather than the original 2.0 mm. That difference could reflect changes in the polymer’s physical arrangement, processing history, or other effects. It does not, by itself, demonstrate chemical degradation.

03Why the first experiment cannot prove that the backbone stayed intact

Suppose the polymer becomes rigid again after cooling.

It is tempting to conclude that no chemical bonds could have broken. After all, the material seems almost back to normal.

But a bulk property returning towards its original value is not conclusive evidence about molecular structure.

Consider two possible explanations.

Explanation A: Physical softening

The polymer chains remain chemically intact. Heating increases their mobility, and cooling reduces it.

Some rearrangement might remain after cooling, so the original shape or stiffness need not return perfectly.

Explanation B: Physical softening and some chemical degradation

Heating increases chain mobility, but a proportion of backbone bonds also break.

When the sample cools, the remaining chains can still become less mobile and the material can stiffen again. The chemical changes may not be obvious from a simple bending test.

Both explanations could produce a material that softens when heated and hardens when cooled.

The original experiment cannot reliably distinguish them.

Revise the experiment: use an unloaded control

Prepare two identical strips from the same material, with the same thermal and processing histories.

Heat both under identical conditions.

• Strip A carries the test weight.
• Strip B has no external weight.

Measure the lengths, resting shapes, and deformations of both during heating and after cooling.

Why add Strip B?

Because heating can change a polymer’s shape even without an applied load. For example, polymer chains that were previously aligned during manufacturing may rearrange when heated, allowing a stretched film to shrink.

If Strip A shortens during heating, the unloaded control helps determine whether this comes from the applied force or from the release of its earlier processing-induced alignment.

Without that control, the interpretation could be completely wrong.

Add a measurement of molecular integrity

To investigate whether the backbone has actually changed, we need evidence about the molecules themselves.

One possible laboratory technique is size-exclusion chromatography, which separates suitable dissolved polymer molecules according to their behaviour in a size-separation system. With appropriate calibration and analysis, it can provide information about a polymer’s molecular mass distribution.

This is an extension of the core HSC bonding explanation; the detailed analytical technique is not needed to understand the result.

If the sample’s average molecular mass decreases substantially after heating, that may support chain scission: the breaking of covalent bonds along polymer backbones.

If the molecular mass distribution remains essentially unchanged, that is consistent with physical softening without substantial scission.

There are limits. The measurement has finite sensitivity, and other chemical or physical changes can complicate the interpretation. A single test rarely proves that absolutely no bonds have broken.

A particularly useful lesson is that mass loss and backbone breaking are not the same measurement.

04A polymer can lose backbone bonds without losing measurable mass

Imagine one long polymer molecule being cut into two shorter molecules.

The number of polymer molecules increases, but the atoms need not leave the sample.

Depending on the reactions involved, the total mass could stay effectively unchanged even though the backbone structure has changed substantially.

05Softening, glass transition, and melting are not identical

There is another distinction worth making before interpreting experimental results.

An amorphous region of a polymer has no long-range ordered arrangement of its chains.

At sufficiently low temperatures, molecular rearrangement in that region is strongly restricted. The material may behave as a hard, glassy solid.

As its temperature increases through the glass transition region, coordinated movement of chain segments becomes much easier.

The characteristic temperature associated with this change is called the glass transition temperature, \(T_g\).

Above \(T_g\), the polymer may become more flexible, rubbery, or susceptible to deformation.

But that doesn’t necessarily mean it has melted.

What about crystalline regions?

Some polymers are semicrystalline: they contain both amorphous regions and regions where chain segments are arranged in an ordered pattern.

A semicrystalline polymer can therefore have two important temperature-dependent processes:

• Glass transition: increased segmental mobility in its amorphous regions.
• Melting: loss of the ordered arrangement in its crystalline regions.

The melting temperature is represented by \(T_m\).

Between \(T_g\) and \(T_m\), amorphous regions can become more mobile while crystalline regions still restrict larger-scale movement.

At sufficiently high temperatures, the crystalline regions can melt, allowing much greater rearrangement and potentially viscous flow in a thermoplastic.

An entirely amorphous thermoplastic does not have a crystalline melting point in the same sense. It can soften through its glass transition and eventually flow without a distinct crystalline melting event.

Neither glass transition nor ordinary thermoplastic melting requires breaking the covalent backbone.

A contrasting case: a crosslinked polymer

Now compare a conventional thermoplastic with a cured thermoset.

A thermoplastic consists of separate polymer molecules, which may be linear or branched. When sufficiently heated, the molecules can rearrange and flow relative to one another.

A thermoset has covalent crosslinks connecting chains into a network.

Heating a crosslinked polymer can still increase movement within the sections of chain between crosslinks.

That means a thermoset can become more flexible without breaking its backbone.

However, an intact, permanently crosslinked network cannot flow as a collection of separate polymer molecules.

This leads to a result students sometimes overlook:

A polymer can soften without melting, and it can soften without becoming permanently remouldable.

If a permanently crosslinked material is heated severely enough, chemical degradation may eventually occur instead of ordinary thermoplastic flow.

06Keep molecular structure, processing, and bulk properties separate

Many polymer questions become confusing because three different levels of explanation get mixed together.

These levels interact, but they are not interchangeable.

Processing can change chain orientation and crystallinity without changing the chemical bonds in the backbone. Some processing conditions can also cause chemical reactions, such as curing or degradation, so the distinction must be tested rather than assumed.

For example, a stretched polymer film may shrink when warmed as its previously aligned chains relax.

The film can then become rigid again after cooling, even though it has not returned to its original length.

Its shape change may be permanent while the temperature-dependent reduction in stiffness is largely reversible.

That is not a contradiction. Shape and stiffness are different properties.

A practical way to interpret a softening experiment

This is the useful experimental decision: identify what was measured first, then decide which molecular explanation the evidence can actually support.

07Questions and solutions

All numerical datasets below are illustrative. Unless stated otherwise, assume the polymers are tested in controlled conditions and that their chemical structures do not change during ordinary softening.

Question 1

A thermoplastic strip bends downwards by 2.0 mm at 25°C when a force of 0.20 N is applied.

When the same strip is heated to 70°C, it bends downwards by 9.0 mm under the same force and loading time.

Assume the polymer undergoes no chemical reaction during the test.

(a) Calculate how many times greater the deflection is at 70°C.

(b) Explain why this change can occur without breaking the polymer’s covalent backbone.

Solution 1

The strip bends 4.5 times further at 70°C because increased molecular mobility makes it easier to deform under the same load.

(a) Calculate the ratio.

\[
\begin{aligned}
\text{Deflection ratio}
&=\frac{9.0\ \mathrm{mm}}{2.0\ \mathrm{mm}}\\
&=4.5
\end{aligned}
\]

The deflection is therefore 4.5 times the original value.

(b) Explain the molecular change.

Heating increases the ability of polymer chain segments to move and rearrange.

The chains can change conformation and, depending on the polymer structure, move relative to neighbouring chains.

This increases deformation under the applied force without requiring the covalent bonds connecting the atoms within each chain to break.

The tempting mistake is to assume that because the strip became easier to bend, its carbon-carbon bonds must have weakened or broken.

The measured change concerns the mechanical response of the bulk material, not the strength of an individual covalent bond.

Question 2

Two strips, A and B, are manufactured from the same hypothetical semicrystalline polymer.

Their chemical structures, molecular mass distributions, dimensions, and test conditions are identical, but their cooling histories differ.

Strip A was cooled rapidly after processing. Strip B was cooled slowly.

Further analysis gives the following illustrative results:

Both strips are tested under the same load for the same time.

The polymer has a glass transition temperature of 50°C and a crystalline melting temperature of 125°C.

(a) Explain why Strip A can bend further even though its covalent backbone is chemically identical to Strip B’s.

(b) Has the crystalline part of either strip melted at 85°C? Explain.

(c) The strips are subsequently heated above 125°C without chemical degradation, then cooled using identical conditions. Predict what should happen to the original difference in their crystalline fractions, and explain any limitation of this prediction.

Solution 2

Strip A bends further because its lower crystalline fraction leaves fewer ordered regions restricting deformation. Neither sample has reached its stated crystalline melting temperature, and identical remelting and cooling conditions should substantially reduce the difference caused by their previous cooling histories.

(a) Explain the different deflections.

Strip A contains 15% crystalline material, while Strip B contains 45%.

The crystalline regions contain more ordered arrangements of polymer chains. These regions can act as physical constraints that limit movement.

At 85°C, the amorphous portions are above the stated glass transition temperature of 50°C. Their chain segments are therefore relatively mobile.

Strip A has a larger amorphous fraction and fewer crystalline constraints. This is consistent with its greater observed deflection.

No difference in backbone covalent bond strength is needed.

(b) Identify the thermal state.

The test temperature is 85°C, which is below the stated crystalline melting temperature of 125°C.

Therefore, the crystalline regions have not undergone their main melting transition.

The amorphous regions can still be mobile above the glass transition temperature.

This demonstrates why softening and crystalline melting should not be treated as identical.

(c) Predict the effect of identical reprocessing.

Heating above the melting temperature removes the original crystalline ordering, provided melting is completed.

If both samples then experience the same controlled cooling conditions, their resulting crystallinity would generally be expected to become more similar.

The earlier difference arose from their different cooling histories, rather than different chemical backbones.

However, identical cooling does not guarantee perfectly identical material properties. Differences in orientation, prior physical history, and experimental conditions can still matter. The prediction assumes those influences are also adequately controlled.

The key trap is assuming that identical chemical composition always guarantees identical mechanical properties.

Question 3

A technician tests a single polymer strip using different loads at different temperatures.

Every measurement is taken after 30 seconds. Assume that deflection is approximately proportional to applied force over the small force range used.

The results are:

Define the apparent bending compliance \(C\) as deflection divided by force:

\[
C=\frac{\delta}{F}
\]

Express compliance in \(\mathrm{mm\,N^{-1}}\).

A technician makes two claims:

• “The polymer became only about 2.7 times more compliant when heated because the deflection increased from 3.0 mm to 8.0 mm.”
• “The cooled strip bends further than it did originally, so some covalent backbone bonds must have broken.”

(a) Calculate the compliance for all three conditions.

(b) Evaluate both claims.

(c) Suggest one additional experiment or measurement that would help distinguish a physical change in the polymer’s arrangement from chemical chain scission.

Solution 3

The heated strip has eight times the initial apparent compliance, not 2.7 times. The changed response after cooling does not, by itself, prove covalent bond cleavage.

(a) Calculate compliance.

Initially:

\[
\begin{aligned}
C_{25}
&=\frac{3.0}{0.30}\\
&=10.0\ \mathrm{mm\,N^{-1}}
\end{aligned}
\]

When heated:

\[
\begin{aligned}
C_{75}
&=\frac{8.0}{0.10}\\
&=80.0\ \mathrm{mm\,N^{-1}}
\end{aligned}
\]

After cooling:

\[
\begin{aligned}
C_{\mathrm{cooled}}
&=\frac{4.0}{0.30}\\
&=13.3\ \mathrm{mm\,N^{-1}}
\end{aligned}
\]

The increase in compliance on heating is

\[
\begin{aligned}
\frac{C_{75}}{C_{25}}
&=\frac{80.0}{10.0}\\
&=8.0
\end{aligned}
\]

Thus the heated strip has eight times the initial apparent bending compliance.

(b) Evaluate the claims.

The first claim ignores the different applied forces.

The heated sample bends further despite being subjected to only one-third of the original force.

Because the load changed, comparing deflection values alone is misleading. Under the assumed approximately proportional response, compliance provides the fairer comparison.

The second claim also goes beyond the available evidence.

After cooling, the compliance is \(13.3\ \mathrm{mm\,N^{-1}}\), compared with \(10.0\ \mathrm{mm\,N^{-1}}\) originally.

The percentage increase is

\[
\begin{aligned}
\%\text{ increase}
&=\frac{13.3-10.0}{10.0}\times100\\
&\approx33\%
\end{aligned}
\]

The material has not returned to its original bending response.

However, this could result from changes in chain orientation, crystallinity, residual deformation, or chemical changes. The mechanical test does not distinguish these explanations.

(c) Suggest an improved test.

One useful improvement would be to compare the polymer’s molecular mass distribution before and after heating.

A substantial shift towards shorter molecules would support the possibility of chain scission, provided the analytical method and other possible changes are considered.

An unloaded control would also help identify changes in shape caused by thermal relaxation rather than by the applied force.

Neither the remaining deformation nor the return of most of the original stiffness is a direct measurement of covalent bond integrity.

Question 4

Two polymers are heated from 25°C to 85°C.

Material X is an amorphous thermoplastic made from separate, linear polymer molecules.

Material Y is a permanently crosslinked polymer network.

Both have a glass transition temperature of approximately 65°C, and both are chemically stable at 85°C.

The following observations are made:

• Both materials become easier to deform at 85°C.
• Under sustained pressure, X flows sufficiently to fill a mould cavity and can retain the moulded shape after cooling.
• Y becomes more flexible but remains a connected network and does not undergo sustained flow. In the stated test, it substantially recovers its original shape when the pressure is removed.

A student concludes:

“X must have broken its backbone bonds because it flowed. Y’s backbone bonds cannot have become more flexible because it did not flow.”

(a) Explain why both parts of the conclusion are incorrect.

(b) Explain why having similar glass transition temperatures does not mean the two polymers must have identical processing behaviour.

(c) Is it scientifically precise to describe X as having passed through a crystalline melting point at 85°C? Justify your answer.

Solution 4

Neither material needs to break its covalent backbone to become softer, and the difference in their flow behaviour is explained by the presence or absence of permanent covalent crosslinks.

(a) Explain the incorrect conclusions.

Material X contains separate linear polymer molecules.

At 85°C, which is above its stated glass transition temperature, chain segments have greater mobility. Under sustained pressure, the molecules can rearrange and eventually move past one another.

This can produce viscous flow without requiring the covalent backbones to break.

Material Y also contains chain segments capable of increased movement at 85°C.

However, its permanent covalent crosslinks prevent unrestricted movement of entire chains relative to one another.

Sections between crosslinks can still change conformation, so the network may become more flexible.

The crosslinks limit whole-chain flow, not all molecular movement.

(b) Explain why equal glass transition temperatures are insufficient.

Glass transition describes a change in molecular mobility, particularly in amorphous regions.

It does not describe every constraint on movement.

For X, the separate polymer molecules can eventually undergo large-scale rearrangement.

For Y, the covalent network remains connected. Large-scale rearrangement is restricted by those permanent links.

Consequently, similar glass transition temperatures do not imply similar ability to flow, retain a moulded shape, or undergo permanent deformation.

The tempting mistake is treating the glass transition as a universal temperature at which every polymer becomes a freely flowing liquid.

(c) Distinguish softening from crystalline melting.

No. Material X is specified to be amorphous, so it does not have a crystalline melting point of the kind associated with a semicrystalline polymer.

It can become increasingly mobile above its glass transition and eventually flow.

Describing it as having softened and undergone viscous flow is more precise than claiming that its crystalline regions melted, because it has no such regions in the stated model.

Question 5

A manufacturer investigates two films made from the same linear, uncrosslinked polymer.

Film A was stretched during manufacture and cooled while its chains were highly oriented.

Film B was previously heat-treated to relax much of that orientation, then cut to the required dimensions.

Both have identical chemical compositions and initial molecular mass distributions.

Each test specimen begins with a measured gauge length of 100 mm at 25°C.

Separate, equivalent specimens are heated to 80°C for 60 seconds under two conditions:

• No applied tensile force.
• A constant tensile force of 0.20 N.

The temperature is high enough to allow substantial chain movement. The specimens do not melt into freely flowing liquids during the test.

The measured lengths at 80°C are:

Assume the specimens have equal measured cross-sectional areas at the test temperature and that the applied tensile force acts uniformly along their gauge lengths.

A technician argues:

“Film A became shorter even though it was being pulled. Therefore, it must resist tensile deformation more strongly than Film B, which became longer. The results also suggest that Film A’s covalent bonds must have changed.”

(a) Determine the effect of the tensile force on the length of each film, using the correct control.

(b) Calculate the load-induced percentage extension relative to each film’s unloaded length at 80°C. Which film has the greater fractional response to the same tensile force?

(c) Explain how Film A can become shorter overall under tensile load while still being extended by that load.

(d) After Film A is heated without a load and cooled again, it remains 86 mm long but becomes comparatively rigid. Explain why this does not contradict the physical-softening model.

(e) Suppose neither film loses measurable mass during heating. Does that prove their covalent backbones stayed intact? Identify what further evidence would be useful.

Solution 5

Film A does not contract because of the tensile force. It extends by 6 mm relative to its unloaded control, just as Film B does. Its overall shortening occurs because relaxation of its manufacturing-induced orientation outweighs the extension caused by the load.

The more subtle result is that equal absolute extensions do not give equal fractional extensions when the reference lengths differ.

(a) Isolate the effect of the applied force.

For Film A:

\[
\begin{aligned}
\Delta L_A
&=92-86\\
&=+6\ \mathrm{mm}
\end{aligned}
\]

For Film B:

\[
\begin{aligned}
\Delta L_B
&=106-100\\
&=+6\ \mathrm{mm}
\end{aligned}
\]

Here, \(\Delta L\) is the load-induced change in length, measured relative to the corresponding unloaded specimen at the same temperature and time.

A positive value means extension.

Both films extend by 6 mm under the applied force.

The initial 100 mm length is not the correct unloaded reference for isolating the effect of the force at 80°C, because Film A also changes length due to heating.

(b) Compare the fractional extensions.

The appropriate reference lengths are 86 mm for A and 100 mm for B.

For Film A:

\[
\begin{aligned}
\varepsilon_A
&=\frac{6}{86}\\
&=0.0698\\
&=6.98\%
\end{aligned}
\]

For Film B:

\[
\begin{aligned}
\varepsilon_B
&=\frac{6}{100}\\
&=0.0600\\
&=6.00\%
\end{aligned}
\]

The symbol \(\varepsilon\) represents engineering strain, calculated here as load-induced extension divided by the unloaded gauge length at the test temperature.

Film A has the greater fractional extension.

Because the applied forces and hot cross-sectional areas are equal, the films are subjected to equal nominal tensile stresses.

Under the stated conditions, Film A therefore shows a slightly greater fractional response to the applied force.

However, this is a time-dependent deformation test. It does not automatically give a unique, time-independent Young’s modulus.

There are two traps here.

The first is comparing the final heated lengths with the original 100 mm lengths without considering the unloaded controls.

The second is correctly finding that both extend by 6 mm, then assuming their fractional deformation must also be identical.

(c) Explain the competing length changes.

Film A was previously stretched and cooled while its polymer chains were highly oriented.

When heated, those chains can become more mobile and relax towards less oriented arrangements.

This can cause the film to shrink along the direction in which it was originally stretched.

Without an applied load, Film A shrinks from 100 mm to 86 mm.

Under the tensile load, the corresponding film reaches 92 mm.

The load therefore opposes some of the thermal shrinkage, even though the final length is still smaller than the original length.

The net change from the original length is

\[
\begin{aligned}
\Delta L_{\mathrm{net},A}
&=92-100\\
&=-8\ \mathrm{mm}
\end{aligned}
\]

The negative sign means that the final length is 8 mm shorter than the original gauge length.

It does not mean the tensile force produced contraction.

The unloaded control reveals that the heating-induced shortening was greater than the load-induced extension.

Film B had already undergone much of this structural relaxation before the experiment, so it shows little net thermal shrinkage in the unloaded condition.

No difference in the chemical strength of their covalent backbones is required to explain these observations.

(d) Explain the permanent change in length but recovery of rigidity.

Film A can remain shorter after cooling because its polymer chains have undergone a rearrangement that is not fully reversed.

Cooling reduces their mobility, making further rearrangement more difficult.

The shorter arrangement may therefore become effectively fixed at the lower temperature.

This produces two different outcomes:

• The physical shape has changed and does not return to its original dimensions.
• The mechanical response becomes more rigid again as molecular movement becomes restricted.

Permanent shape change is not the same as permanent chemical change.

A polymer can be reshaped physically while its covalent backbone remains intact.

(e) Evaluate the evidence from mass measurements.

No measurable mass loss does not prove that no covalent bonds broke.

A backbone bond could break and create two shorter molecules while both fragments remain in the material.

The total mass could remain essentially unchanged.

To investigate the molecular structure, the manufacturer could compare molecular mass distributions before and after heating using an appropriate analytical method.

A significant shift towards shorter chains would support chemical chain scission, although other possible chemical changes and measurement limitations would need consideration.

Conversely, an essentially unchanged distribution, together with controlled heating below the established degradation range, would strengthen the explanation based on physical chain movement and processing-induced relaxation.

The central conclusion is that Film A’s behaviour must be interpreted using its unloaded control, its processing history, and an appropriate molecular measurement. Final length alone cannot establish stiffness or bond breaking.

08What this lets you decide next

When a polymer becomes softer, the first useful question is not whether its strong covalent bonds have become weak. It is whether the temperature has changed how its existing chains can move.

The next question is whether anything chemically irreversible happened during that heating.

A bending test tells you about bulk deformation. An unloaded control reveals changes caused by processing history. Molecular analysis can provide evidence about the integrity of the polymer chains.

Keeping those measurements separate is the foundation for deciding whether a heated polymer has simply softened, become permanently reshaped, or undergone chemical damage.

For the next distinction, see How to Distinguish Polymer Degradation from Melting.