NDA Physics · Heat and Thermodynamics

Temperature, Scales, and Thermal Expansion

Temperature measures the average kinetic energy of a body's molecules; we read it on the Celsius, Fahrenheit, or Kelvin scales, convert between them with two linear formulas, and watch solids and liquids expand as they get hotter.

All Heat and Thermodynamics notes NDA Physics strategy

Why this matters

Start here — every later movement rests on these basics. Scale conversion is the single most-tested skill in the chapter (Celsius to Fahrenheit, Celsius to Kelvin, and the famous 'when do two scales read the same?' problems). Absolute zero (0 K = −273.15°C) is a recurring one-mark recall. Thermal expansion adds a small numeric strand — the linear / areal / volume coefficients are related by a fixed ratio, and the pendulum-slows-when-heated idea shows up. About 8 PYQs, mostly EASY and MODERATE with one HARD scale problem.

Concept 1 of 5

Temperature and the three scales

Intuition

Temperature tells you how hot something is — physically, it is a measure of the AVERAGE kinetic energy of the molecules of a body. It is NOT the same as heat: heat is energy flowing because of a temperature difference, temperature is the reading on the thermometer. Three scales are in use: Celsius (water freezes at 0, boils at 100), Fahrenheit (32 and 212), and Kelvin, the SI scale, which starts at absolute zero and has the same step size as Celsius.

Definition

Temperature is a measure of the average kinetic energy of the particles of a substance. The three common scales:

Celsius (°C) — ice point 0°C, steam point 100°C (100 divisions).
Fahrenheit (°F) — ice point 32°F, steam point 212°F (180 divisions).
Kelvin (K) — the SI absolute scale; 0 K is absolute zero. A change of 1 K equals a change of 1°C (same size step); they differ only by the offset 273.15.

Kelvin is never written with a degree sign — it is '300 K', not '300°K'.

Temperature-scale conversions

\frac{C}{5} = \frac{F - 32}{9} = \frac{K - 273.15}{5} \qquad F = \frac{9}{5}C + 32 \qquad K = C + 273.15

Ctemperature in degrees Celsius
Ftemperature in degrees Fahrenheit
Ktemperature in kelvin (absolute)

Worked example

Convert a body temperature of 37°C to the Fahrenheit and Kelvin scales.

Fahrenheit: $F = \frac{9}{5}C + 32 = \frac{9}{5}(37) + 32$ .
$\frac{9}{5}\times 37 = 66.6$ , so $F = 66.6 + 32 = 98.6°\text{F}$ — the familiar normal body temperature.
Kelvin: $K = C + 273.15 = 37 + 273.15 = 310.15\,\text{K}$ (often rounded to 310 K).

Answer:98.6°F and about 310 K.

Practice this conceptself-check · 4 quick reps

Try it yourself

A place reads 113°F on a sunny day. What is this temperature on the Kelvin scale?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Convert 100°C to Fahrenheit.
2.
Convert 27°C to Kelvin.
3.
A body warms from 310 K to 340 K. By how many °C did it rise?
4.
In '°F = X + 1.8 × °C', what is X?

From the bank · past-year question

Example 1Heat and ThermodynamicsEASY

The temperature of a place on one sunny day is 113 in Fahrenheit scale. The Kelvin scale reading of this temperature will be

[Q98 · Sep · 2019]

A temperature CHANGE is the same in K and °C, but a temperature VALUE is not

If the question asks 'increase of 30 K equals how many °C?', the answer is 30°C — because the step size is identical. But '300 K equals how many °C?' is

300 - 273 = 27°\text{C}

, because of the offset. Read whether it asks for a value or a change.

Never write a degree sign with Kelvin

It is '273 K' and '0 K', not '273°K'. The kelvin already IS an absolute scale, so the degree symbol is dropped by convention. An option that writes '°K' is usually the planted wrong one.

Drill 2 more on temperature and the three scales

Concept 2 of 5

Absolute zero and choosing a thermometer

Intuition

There is a lowest possible temperature — absolute zero — where molecular motion is at its theoretical minimum. It sits at 0 K, which is −273.15°C. No temperature can ever be lower. Different thermometers suit different ranges: a mercury thermometer works for everyday temperatures, but for very low temperatures (around −250°C) you need a thermocouple- or gas-based thermometer, because mercury freezes.

Definition

Absolute zero is the lowest possible temperature: 0 K = −273.15°C (often quoted as −273°C). At this point the thermal kinetic energy of particles is at its minimum. Thermometer choice by range:

Liquid-in-glass (mercury / alcohol) — ordinary lab and clinical use.
Thermocouple thermometer — wide range including very low (around −250°C) and very high temperatures; based on the voltage produced at a junction of two metals.
Constant-volume gas thermometer — the most accurate standard, used to define the Kelvin scale.

Worked example

Express absolute zero on the Celsius and Fahrenheit scales.

Absolute zero is 0 K by definition.
Celsius: $C = K - 273.15 = 0 - 273.15 = -273.15°\text{C}$ .
Fahrenheit: $F = \frac{9}{5}C + 32 = \frac{9}{5}(-273.15) + 32 = -459.67°\text{F}$ .

Answer:−273.15°C and about −459.67°F.

Practice this conceptself-check · 4 quick reps

Try it yourself

You need to measure a temperature of about −250°C in a cryogenics lab. Why can't you use an ordinary mercury thermometer, and what would you use instead?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
What is the lowest possible temperature in °C?
2.
What is absolute zero in kelvin?
3.
Which thermometer suits about −250°C?
4.
Which thermometer type defines the most accurate standard scale?

From the bank · past-year question

Example 2Heat and ThermodynamicsEASY

Which one of the following instruments can be used to measure -250 degree C temperature ?

[Q63 · Sep · 2025]

Absolute zero is −273°C, not −273 K

Absolute zero is 0 K, equivalently −273.15°C. A distractor that says 'absolute zero is −273 K' confuses the two scales — −273 K is meaningless because the Kelvin scale cannot go below 0.

Drill 2 more on absolute zero and choosing a thermometer

Concept 3 of 5

When do two scales read the same?

Intuition

A favourite NDA trick: 'at what temperature do the Celsius and Fahrenheit scales show the same number?' or even 'where do Kelvin and Fahrenheit agree?'. The method is always the same — set the two scale variables equal, substitute one conversion into the other, and solve the resulting linear equation.

Definition

To find where two scales read the same numerical value, set their variables equal and solve:

C = F: put $F = C$ into $F = \frac{9}{5}C + 32$ and solve. Answer: −40° (the one temperature where Celsius and Fahrenheit coincide).
K = F: put $K = F$ into $K = C + 273$ and $F = \frac{9}{5}C + 32$ , then eliminate C.

The trick is purely algebraic: two linear relations, one unknown.

Worked example

At what temperature do the Celsius and Fahrenheit thermometers show the same reading?

Set $F = C$ (same numerical value).
Substitute into $F = \frac{9}{5}C + 32$ : $C = \frac{9}{5}C + 32$ .
Move terms: $C - \frac{9}{5}C = 32 \Rightarrow -\frac{4}{5}C = 32$ .
So $C = 32 \times \left(-\frac{5}{4}\right) = -40$ .

Answer:−40° (i.e. −40°C = −40°F).

Practice this conceptself-check · 3 quick reps

Try it yourself

A Kelvin thermometer and a Fahrenheit thermometer give the same numerical reading for one sample. What is the corresponding Celsius reading?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
At what Celsius temperature does the Fahrenheit scale read the same number?
2.
Set up the equation for 'K equals F'.
3.
If C = F, what is that single common reading?

From the bank · past-year question

Example 3Heat and ThermodynamicsHARD

A Kelvin thermometer and a Fahrenheit thermometer both give the same reading for a certain sample. What would be the corresponding reading in a Celsius thermometer ?

[Q121 · Apr · 2017]

Set the SCALE VARIABLES equal, not the formula sides

The condition 'two scales read the same number' means

F = C

(or

K = F

) — set those equal, THEN use a conversion to get one equation in one unknown. Do not just equate

\frac{9}{5}C+32

to something; first decide which two readings coincide.

Drill 1 more on when do two scales read the same?

Concept 4 of 5

Thermal expansion — linear, areal, and volume coefficients

Intuition

Heat a solid and it grows in every dimension. A rod gets longer (linear expansion), a sheet grows in area (areal), and a block grows in volume (volume expansion). The three coefficients are not independent — they are locked in a fixed ratio because area is length-squared and volume is length-cubed.

Definition

For a solid heated through $\Delta\theta$ :

Linear: $\Delta L = L\alpha\,\Delta\theta$ — $\alpha$ is the coefficient of linear expansion.
Areal (superficial): $\Delta A = A\beta\,\Delta\theta$ — $\beta$ is the coefficient of areal expansion.
Volume (cubical): $\Delta V = V\gamma\,\Delta\theta$ — $\gamma$ is the coefficient of volume expansion.

They are related by the fixed ratio $\alpha : \beta : \gamma = 1 : 2 : 3$ , so $\beta = 2\alpha$ and $\gamma = 3\alpha = \tfrac{3}{2}\beta$ . Anomalous expansion of water: water is the famous exception — between 0°C and 4°C it CONTRACTS on heating, reaching maximum density at 4°C. Above 4°C it expands normally. This is why ice floats and ponds freeze from the top down (the diagram below shows the density peak).

Expansion coefficients are in the ratio 1 : 2 : 3

\Delta L = L\alpha\,\Delta\theta \qquad \beta = 2\alpha \qquad \gamma = 3\alpha = \tfrac{3}{2}\beta

\alphalinear expansion coefficient
\betaareal (superficial) expansion coefficient
\gammavolume (cubical) expansion coefficient
\Delta\thetarise in temperature

Water is densest at 4 degrees C: between 0 and 4 degrees C it contracts on heating (anomalous), so colder water and ice float — which is why ponds freeze top-down and fish survive below.

Worked example

The coefficient of areal expansion of a material is

1.6\times 10^{-5}\,\text{K}^{-1}

. Find its coefficient of volume expansion.

Volume relates to areal by $\gamma = \frac{3}{2}\beta$ (since $\gamma = 3\alpha$ and $\beta = 2\alpha$ ).
So $\gamma = \frac{3}{2}\times 1.6\times 10^{-5}$ .
$\gamma = 2.4\times 10^{-5}\,\text{K}^{-1}$ .

Answer:

2.4\times 10^{-5}\,\text{K}^{-1}

Practice this conceptself-check · 3 quick reps

Try it yourself

A metal rod has linear expansion coefficient

\alpha = 2\times 10^{-5}\,\text{K}^{-1}

. What are its areal and volume expansion coefficients?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
If $\alpha = 1.7\times 10^{-5}$ , what is $\gamma$ ?
2.
What is the ratio $\alpha:\beta:\gamma$ ?
3.
A 2 m rod with $\alpha = 10^{-5}\,\text{K}^{-1}$ is heated 50 K. Find $\Delta L$ .

From the bank · past-year question

Example 4Heat and ThermodynamicsMODERATE

The coefficient of areal expansion of a material is

1\cdot6 \times 10^{-5}

^{-1}

. Which one of the following gives the value of coefficient of volume expansion of this material?

[Q112 · Sep · 2018]

Areal to volume is ×3/2, not ×3

Going linear to volume multiplies by 3. But going AREAL to volume multiplies by only

\frac{3}{2}

, because areal is already

2\alpha

. Decide which coefficient you were given before scaling.

Concept 5 of 5

Consequences of expansion — pendulums and liquid measurement

Intuition

Thermal expansion shows up in two classic NDA scenarios. A pendulum clock runs SLOW in summer because its rod lengthens, increasing the time period. And measuring a liquid's expansion is tricky because the container expands too — so you only ever see the APPARENT expansion unless you correct for the vessel.

Definition

Pendulum clock: the period is $T = 2\pi\sqrt{L/g}$ . When the rod is heated, $L$ grows, so $T$ increases — the clock loses time (runs slow) in hot weather. The fractional change is small: $\frac{\Delta T}{T} = \tfrac{1}{2}\alpha\,\Delta\theta$ .
Liquid expansion: a liquid is held in a container that ALSO expands. The observed rise gives only the apparent expansion; the real (absolute) expansion = apparent expansion + expansion of the container. This is why a liquid's coefficient is harder to measure than a solid's.

Pendulum period and apparent expansion

T = 2\pi\sqrt{\frac{L}{g}} \qquad \frac{\Delta T}{T} = \frac{1}{2}\alpha\,\Delta\theta \qquad \gamma_{\text{real}} = \gamma_{\text{apparent}} + \gamma_{\text{vessel}}

Ttime period of the pendulum
Llength of the pendulum rod
\gamma_{\text{real}}true volume expansion of the liquid
\gamma_{\text{apparent}}observed expansion (uncorrected)

Worked example

A pendulum clock made with a copper rod is moved into a hotter room (temperature up by 30°C). What happens to its time period?

The rod expands: $L$ increases by $\Delta L = L\alpha\,\Delta\theta$ .
The period $T = 2\pi\sqrt{L/g}$ grows with $\sqrt{L}$ , so a longer rod gives a longer period.
The fractional rise $\frac{\Delta T}{T} = \frac{1}{2}\alpha\,\Delta\theta$ is tiny (copper $\alpha \approx 17\times 10^{-6}\,\text{K}^{-1}$ ), so the increase is slight.

Answer:The time period increases slightly (the clock runs slow).

Practice this conceptself-check · 3 quick reps

Try it yourself

Why is it harder to measure the coefficient of expansion of a liquid than of a solid?

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
Does a pendulum clock run fast or slow when heated?
2.
Real expansion of a liquid = apparent expansion + ?
3.
Fractional change in pendulum period for a rise $\Delta\theta$ ?

From the bank · past-year question

Example 5Heat and ThermodynamicsMODERATE

The time period of a simple pendulum made using a thin copper wire of length

L

T

. Suppose the temperature of the room in which this simple pendulum is placed increases by 30°C, what will be the effect on the time period of the pendulum ?

[Q107 · Apr · 2017]

Heated pendulum slows DOWN — the period goes UP

Students sometimes say 'faster' assuming heat speeds things up. Physically the rod lengthens, the period

T = 2\pi\sqrt{L/g}

rises, so the clock ticks slower and LOSES time. The effect is slight, not 'more than double'.

Drill 1 more on consequences of expansion — pendulums and liquid measurement

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Temperature and the three scales
Temperature-scale conversions
$\frac{C}{5} = \frac{F - 32}{9} = \frac{K - 273.15}{5} \qquad F = \frac{9}{5}C + 32 \qquad K = C + 273.15$
Thermal expansion — linear, areal, and volume coefficients
Expansion coefficients are in the ratio 1 : 2 : 3
$\Delta L = L\alpha\,\Delta\theta \qquad \beta = 2\alpha \qquad \gamma = 3\alpha = \tfrac{3}{2}\beta$
Consequences of expansion — pendulums and liquid measurement
Pendulum period and apparent expansion
$T = 2\pi\sqrt{\frac{L}{g}} \qquad \frac{\Delta T}{T} = \frac{1}{2}\alpha\,\Delta\theta \qquad \gamma_{\text{real}} = \gamma_{\text{apparent}} + \gamma_{\text{vessel}}$

Watch out for (6)

A temperature CHANGE is the same in K and °C, but a temperature VALUE is not
→ Temperature and the three scales
Never write a degree sign with Kelvin
→ Temperature and the three scales
Absolute zero is −273°C, not −273 K
→ Absolute zero and choosing a thermometer
Set the SCALE VARIABLES equal, not the formula sides
→ When do two scales read the same?
Areal to volume is ×3/2, not ×3
→ Thermal expansion — linear, areal, and volume coefficients
Heated pendulum slows DOWN — the period goes UP
→ Consequences of expansion — pendulums and liquid measurement

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Heat and ThermodynamicsEASY

The formula for conversion between Fahrenheit and Celsius is

°F = X + (1.8 \times °C)

. What is factor

X

[Q117 · Apr · 2019]

Example 2Heat and ThermodynamicsEASY

Which one of the following is the lowest possible temperature ?

[Q63 · Apr · 2021]

Example 3Heat and ThermodynamicsMODERATE

Numerically two thermometers, one in Fahrenheit scale and another in Celsius scale shall read same at

[Q64 · Apr · 2021]

Example 4Heat and ThermodynamicsMODERATE

Why is it difficult to measure the coefficient of expansion of a liquid than solid ?

[Q124 · Apr · 2017]

Example 5Heat and ThermodynamicsEASY

The temperature of a body increases from 310 K to 340 K. The temperature increase in degree Celsius is

[Q55 · Sep · 2025]

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