Temperature — thermal equilibrium, thermometers, scales: Exercises

Level 1 — Recognition

Exercise 1.1

A metal doorknob and a wooden door in the same cold room have been sitting untouched all night. The doorknob feels colder to your hand. Which is actually at the higher temperature?

Recall Solution

Neither — they are at the same temperature. Both objects have been in the same room long enough to reach thermal equilibrium with the air (and each other, by the Zeroth Law of Thermodynamics). "Feels colder" only means the metal pulls heat out of your warm hand faster because it conducts heat well. Feeling is about rate of heat flow, not temperature.

Exercise 1.2

Water freezes and water boils (at 1 atm). Fill in the readings on each scale.

	Ice point	Steam point
Celsius	?	?
Fahrenheit	?	?
Kelvin	?	?

Recall Solution

	Ice point	Steam point
Celsius	$0 ° C$	$100 ° C$
Fahrenheit	$32 ° F$	$212 ° F$
Kelvin	$273.15 K$	$373.15 K$

Why: Celsius and Kelvin both split the ice→steam gap into 100 equal steps, so a degree is the same size in both; Kelvin just shifts the origin down by $273.15$ . Fahrenheit uses 180 steps, so its degree is smaller.

Level 2 — Application

Exercise 2.1

Convert $t_{C} = 20 ° C$ (a comfortable room) to Fahrenheit and to Kelvin.

Recall Solution

Fahrenheit: $t_{F} = \frac{9}{5} (20) + 32 = 36 + 32 = 68 ° F$ . Kelvin: $T = 20 + 273.15 = 293.15 K$ . Why the two steps for Fahrenheit: multiply by $\frac{9}{5}$ to stretch the interval size (F degrees are smaller), then add $32$ to shift the origin (F puts the ice point at $32$ , not $0$ ).

Exercise 2.2

A weather report says it is $t_{F} = 5 ° F$ . Convert to Celsius.

Recall Solution

$t_{C} = \frac{5}{9} (t_{F} - 32) = \frac{5}{9} (5 - 32) = \frac{5}{9} (- 27) = - 15 ° C$ . Why subtract 32 first: we must undo the origin shift before rescaling the interval — the $- 32$ removes Fahrenheit's offset, then $\frac{5}{9}$ shrinks the interval back to Celsius size.

Exercise 2.3

Convert $T = 300 K$ to Celsius and to Fahrenheit.

Recall Solution

Celsius: $t_{C} = 300 - 273.15 = 26.85 ° C$ . Fahrenheit: $t_{F} = \frac{9}{5} (26.85) + 32 = 48.33 + 32 = 80.33 ° F$ . Why go through Celsius: the Fahrenheit formula is written in terms of Celsius, so Kelvin → Celsius → Fahrenheit is the safe two-hop route.

Level 3 — Analysis

Exercise 3.1

On a hot day a room warms from $22 ° C$ to $34 ° C$ . Express this change in Kelvin and in Fahrenheit degrees.

Recall Solution

The Celsius change is $Δ t_{C} = 34 - 22 = 12 ° C$ . In Kelvin: $Δ T = Δ t_{C} = 12 K$ . In Fahrenheit: $Δ t_{F} = \frac{9}{5} (12) = 21.6 ° F$ . Why no $+ 273.15$ or $+ 32$ : an interval is a difference of two readings. Both readings carry the same offset, so it cancels when you subtract. Only the slope ( $1$ for Kelvin, $\frac{9}{5}$ for Fahrenheit) survives.

Exercise 3.2

At what temperature do the Celsius and Fahrenheit scales read the same number?

Recall Solution

Set $t_{F} = t_{C} = x$ : $x = \frac{9}{5} x + 32 \Rightarrow x - \frac{9}{5} x = 32 \Rightarrow - \frac{4}{5} x = 32 \Rightarrow x = - 40.$ So $- 40 ° C = - 40 ° F$ . What it looks like (see figure): two straight lines on a $t_{C}$ – $t_{F}$ plot cross exactly once, at $(- 40, - 40)$ .

Figure — Temperature — thermal equilibrium, thermometers, scales

Exercise 3.3

A faulty Celsius thermometer reads $2 ° C$ in melting ice and $102 ° C$ in boiling water (at 1 atm). It reads $52 ° C$ in some liquid. What is the true Celsius temperature?

Recall Solution

The faulty scale is still linear, so the fraction of the way from ice to steam must be the same on both scales: $\frac{t _{true} - 0}{100 - 0} = \frac{t _{faulty} - 2}{102 - 2} .$ Plug in $t_{faulty} = 52$ : $t_{true} = 100 \cdot \frac{52 - 2}{100} = 100 \cdot \frac{50}{100} = 50 ° C .$ Why the ratio method: any two linear scales agree on fractions between their fixed points, so equating those fractions converts one to the other without ever needing $a$ and $b$ separately.

Level 4 — Synthesis

Exercise 4.1

A constant-volume gas thermometer reads pressure $P_{tr} = 4.0 \times 1 0^{4} Pa$ at the triple point of water ( $T_{tr} = 273.16 K$ ). In a hot oil bath it reads $P = 5.5 \times 1 0^{4} Pa$ . Find the bath temperature in Kelvin and in Celsius.

Recall Solution

At constant volume the absolute (single-fixed-point) definition gives temperature proportional to pressure: $T = T_{tr} \frac{P}{P _{tr}} = 273.16 \times \frac{5.5 \times 1 0 ^{4}}{4.0 \times 1 0 ^{4}} = 273.16 \times 1.375 = 375.60 K .$ In Celsius: $t_{C} = 375.60 - 273.15 = 102.45 ° C$ . Why just scale the pressure: the Ideal Gas Law at fixed volume makes $P \propto T$ (absolute), so the pressure ratio is the temperature ratio. This is why the gas thermometer defines the true Kelvin scale.

Exercise 4.2

The pressure of a fixed volume of dilute gas obeys $P = P_{0} (1 + α t_{C})$ with $α = \frac{1}{273.15} ° C^{- 1}$ . Show that extrapolating to $P = 0$ gives absolute zero, and state its value.

Recall Solution

Set $P = 0$ : $0 = P_{0} (1 + α t_{C}) \Rightarrow 1 + α t_{C} = 0 \Rightarrow t_{C} = - \frac{1}{α} = - 273.15 ° C .$ What it looks like (see figure): the straight line of $P$ vs $t_{C}$ , extended backwards through the axis, hits zero pressure at $- 273.15 ° C$ . Since a real gas cannot push with negative pressure, no temperature can lie below this — it is absolute zero $= 0 K$ .

Level 5 — Mastery

Exercise 5.1

Two gas thermometers, X (constant volume) and a liquid thermometer Y, are both calibrated to agree at $0 ° C$ and $100 ° C$ . At the true temperature $50 ° C$ , thermometer Y reads $48 ° C$ because its liquid does not expand perfectly linearly. Meanwhile the gas thermometer X reads $50 ° C$ . (a) Which thermometer defines the "true" absolute scale, and why? (b) By how much does Y disagree with X at this point?

Recall Solution

(a) The gas thermometer X defines the true scale. All dilute gases follow the same universal $P$ – $T$ line (that's the kinetic-theory result behind the Ideal Gas Law), so they all agree with each other. A liquid like Y is defined to be linear only at its two fixed points; between them its real expansion curves slightly, so it drifts. (b) Disagreement $= t_{X} - t_{Y} = 50 - 48 = 2 ° C$ . Y reads $2 ° C$ too low at the midpoint. Why gases win: at low pressure the molecules barely interact, so behaviour becomes substance-independent — the definition of an absolute standard.

Exercise 5.2

A steel bridge span is $t_{C} = 15 ° C$ at installation. In summer it reaches $45 ° C$ ; in winter, $- 10 ° C$ . (a) Give the summer and winter changes from installation in Kelvin. (b) These changes feed the thermal-expansion formula $Δ L = L_{0} α_{L} Δ T$ . Explain why we may use $Δ T$ in Kelvin or $Δ t_{C}$ interchangeably here.

Recall Solution

(a) Summer change: $Δ t_{C} = 45 - 15 = 30 ° C \Rightarrow Δ T = 30 K$ . Winter change: $Δ t_{C} = - 10 - 15 = - 25 ° C \Rightarrow Δ T = - 25 K$ (the span contracts). (b) Because a Kelvin degree and a Celsius degree are the same size (both split ice→steam into 100 parts). Only the origin differs by $273.15$ , and an origin shift cancels in any difference. So $Δ T = Δ t_{C}$ exactly — the Thermal Expansion formula gives identical numbers with either. Why this matters: if you carelessly plugged the absolute temperature $T = 318 K$ instead of the change $Δ T = 30 K$ , you'd overestimate the expansion tenfold.

Recall One-line summary of every tool used

Convert readings ::: $t_{F} = \frac{9}{5} t_{C} + 32$ , $T = t_{C} + 273.15$ Convert changes ::: $Δ t_{F} = \frac{9}{5} Δ t_{C}$ , $Δ T = Δ t_{C}$ (offsets cancel) Compare linear thermometers ::: equate fractions $\frac{t - ice}{steam - ice}$ Gas thermometer ::: $T = T_{tr} P / P_{tr}$ at constant volume Absolute zero ::: extrapolate $P = P_{0} (1 + α t_{C})$ to $P = 0 \Rightarrow - 273.15 ° C$