Temperature — thermal equilibrium, thermometers, scales: Worked examples

This is a companion drill page for the parent topic. The parent built the ideas; here we hunt down every kind of question temperature conversions and thermometers can throw at you — every sign, every degenerate input, every limiting case — and work each one from zero.

The scenario matrix

Every temperature problem in this chapter is one of these case-classes. The six examples below hit all of them.

Cell	Case class	What makes it tricky	Example
A	Positive value, C→F	plain slope + shift	Ex 1
B	Negative value / below zero	signs survive the $\times \frac{9}{5}$	Ex 2
C	A difference (interval), not a reading	offsets ( $+ 32$ , $+ 273.15$ ) must cancel	Ex 3
D	Degenerate / crossing point	two lines meet — one equation	Ex 4
E	Gas thermometer, absolute scale	ratio, not shift; limiting $P \to 0$	Ex 5
F	Real-world word problem + exam twist	pick reading vs. difference correctly	Ex 6

Case A — a positive reading, Celsius → Fahrenheit

Worked example Ex 1 · Cell A

A pleasant room is at $t_{C} = 20 ° C$ . What is this on the Fahrenheit scale?

Forecast: Bigger than 32 (above ice point), and each Celsius degree is worth $\frac{9}{5}$ of a Fahrenheit degree, so guess "somewhere in the high 60s."

Step 1 — Reading or Rise? This is a reading (a where), so we use the full formula. Why this step? The room sits at a fixed spot on the scale; we are not measuring a change. Offsets stay in.

Step 2 — Scale the interval size. $\frac{9}{5} \times 20 = 36$ Why this step? The $\frac{9}{5}$ converts the size of the gap above ice point from Celsius degrees to the smaller Fahrenheit degrees.

Step 3 — Shift the origin. $t_{F} = 36 + 32 = 68 ° F$ Why this step? Fahrenheit's ice point is $32$ , not $0$ , so we add $32$ to move from "degrees above ice" to an actual Fahrenheit reading.

Verify: Reverse it: $t_{C} = \frac{5}{9} (68 - 32) = \frac{5}{9} (36) = 20 ° C$ . ✓ Round-trip lands exactly back.

Case B — a negative reading, signs must survive

Case C — a temperature difference (this is where people trip)

Worked example Ex 3 · Cell C

During an experiment a metal block cools by $Δ t_{C} = 40 ° C$ . Express this change in Kelvin and in Fahrenheit degrees.

Forecast: In Kelvin a change equals the Celsius change exactly — so $40$ K. In Fahrenheit it should be bigger (F degrees are smaller), so guess $\approx 72$ .

Step 1 — Reading or Rise? This is a Rise (the word "by" = a change). Why this step? We are told how far the temperature moved, not where it landed. Offsets must cancel.

Step 2 — Kelvin change. $Δ T = Δ t_{C} = 40 K$ Why this step? Kelvin and Celsius have identical degree size; the $+ 273.15$ is a fixed origin shift that subtracts away when you take a difference: $(T_{2} - T_{1}) = (t_{2} + 273.15) - (t_{1} + 273.15)$ .

Step 3 — Fahrenheit change. $Δ t_{F} = \frac{9}{5} Δ t_{C} = \frac{9}{5} (40) = 72 ° F$ Why this step? Only the slope $\frac{9}{5}$ survives a difference; the $+ 32$ cancels the same way. Never add $32$ to a change.

Verify (do it the long way to prove the offset cancels): Say the block went $t_{C} : 90 \to 50$ . In F: $90 \to \frac{9}{5} (90) + 32 = 194$ ; $50 \to \frac{9}{5} (50) + 32 = 122$ . Difference $194 - 122 = 72$ . ✓ The $+ 32$ vanished on subtraction.

Case D — the degenerate "same reading" case

Worked example Ex 4 · Cell D

Is there any temperature where the Celsius and Fahrenheit thermometers show the same number? Find it.

Forecast: The two lines have different slopes ( $1$ vs $\frac{9}{5}$ ) so they cross exactly once — somewhere in the negatives, because at $0 ° C$ F already reads $32$ (F is ahead), so they must have been equal earlier, below zero.

Step 1 — Write "same reading" as one equation. Let the common value be $x$ : $x = \frac{9}{5} x + 32$ Why this step? "Same number on both scales" means substitute $t_{F} = t_{C} = x$ into the conversion — one unknown, one equation, one crossing point of two straight lines.

Step 2 — Collect the $x$ terms. $x - \frac{9}{5} x = 32 \Rightarrow - \frac{4}{5} x = 32$ Why this step? Bringing both $x$ 's to one side isolates the single unknown.

Step 3 — Solve. $x = 32 \times (- \frac{5}{4}) = - 40$ Why this step? Divide by $- \frac{4}{5}$ (i.e. multiply by $- \frac{5}{4}$ ) to release $x$ .

Verify: $\frac{9}{5} (- 40) + 32 = - 72 + 32 = - 40$ . ✓ So $- 40 ° C = - 40 ° F$ — the one place the scales agree.

Figure — Temperature — thermal equilibrium, thermometers, scales

The figure shows the two conversion lines crossing at exactly one point — the red dot at $- 40$ . Above it Fahrenheit reads higher; below it (colder) Celsius reads higher.

Case E — gas thermometer, absolute scale, and the limit $P \to 0$

Worked example Ex 5 · Cell E

A constant-volume gas thermometer reads pressure $P_{tr} = 3.0 \times 1 0^{4} Pa$ at the triple point of water ( $T_{tr} = 273.16 K$ ). In a hot bath it reads $P = 4.5 \times 1 0^{4} Pa$ . (a) Find the bath temperature. (b) What pressure would this thermometer read at absolute zero?

Forecast: Pressure went up by a factor $1.5$ , so temperature should too: $\approx 410$ K. And absolute zero must correspond to $P = 0$ — that's the whole meaning of "absolute."

Step 1 — Use the single-fixed-point absolute law. At constant volume, $T = T_{tr} \frac{P}{P _{tr}}$ Why this step? For a dilute gas at fixed volume, pressure is directly proportional to absolute temperature. One fixed point fixes the whole line because the line must pass through the origin. This is why the gas thermometer defines the absolute scale — no shift, just a ratio.

Step 2 — Plug in (a). $T = 273.16 \times \frac{4.5 \times 1 0 ^{4}}{3.0 \times 1 0 ^{4}} = 273.16 \times 1.5 = 409.74 K$ Why this step? The pressure ratio is $1.5$ , and temperature scales identically because the relation is a pure proportion.

Step 3 — Answer (b), the limit. Set $T \to 0$ : $0 = T_{tr} \frac{P}{P _{tr}} \Rightarrow P = 0 Pa$ Why this step? Absolute zero is defined as the temperature where the extrapolated gas pressure hits zero. A gas cannot push with negative pressure, so this is the hard floor — you cannot go colder.

Verify (units + sanity): $K \times \frac{Pa}{Pa} = K$ ✓ (pressure ratio is dimensionless). And $409.74 K > 273.16 K$ , correct since the bath is hotter than the triple point. ✓

The figure plots pressure against absolute temperature: a straight line through the origin. Extending it back to $P = 0$ lands exactly at $T = 0$ — that intercept is absolute zero.

Case F — real-world word problem with an exam twist

Worked example Ex 6 · Cell F

A weather report says today's high is $95 ° F$ and tonight's low is $59 ° F$ . (a) Convert the high to Celsius. (b) A physics exam asks: "By how many Celsius degrees does the temperature drop overnight?" A student converts each reading and subtracts. Show the shortcut and confirm it matches.

Forecast: $95 ° F$ is hot — around the mid-30s in Celsius. The drop is $95 - 59 = 36$ F-degrees; since C-degrees are bigger, the Celsius drop should be smaller than 36 — guess $\approx 20$ .

Step 1 — Convert the high (a Reading). $t_{C} = \frac{5}{9} (95 - 32) = \frac{5}{9} (63) = 35 ° C$ Why this step? A reading uses the full inverse formula: subtract the $32$ origin, then scale down by $\frac{5}{9}$ to Celsius-sized degrees.

Step 2 — The overnight drop is a Rise (change), so use the slope only. $Δ t_{F} = 95 - 59 = 36 ° F, Δ t_{C} = \frac{5}{9} Δ t_{F} = \frac{5}{9} (36) = 20 ° C$ Why this step? This is the exam twist: a difference, so the $+ 32$ / $- 32$ cancels. Convert the interval with $\frac{5}{9}$ alone.

Step 3 — Confirm with the long way (convert both, subtract). $59 ° F \to \frac{5}{9} (59 - 32) = \frac{5}{9} (27) = 15 ° C; 35 - 15 = 20 ° C$ Why this step? Converting each reading then subtracting must give the same answer as the shortcut — this is the whole point of "offsets cancel."

Verify: Shortcut $= 20 ° C$ , long way $= 20 ° C$ . ✓ And the high converts back: $\frac{9}{5} (35) + 32 = 63 + 32 = 95 ° F$ . ✓

Quick self-test

Recall Which formula for a

change, and why? For a Celsius→Fahrenheit change, use only $Δ t_{F} = \frac{9}{5} Δ t_{C}$ ::: because the $+ 32$ is an origin shift that cancels when you subtract two readings.

Recall Where do Celsius and Fahrenheit read the same number?

At $- 40$ ::: solving $x = \frac{9}{5} x + 32$ gives $x = - 40$ , so $- 40 ° C = - 40 ° F$ .

Recall Why does the gas thermometer use a ratio, not a shift?

Because $P \propto T$ at constant volume with a line through the origin ::: so a single fixed point sets the whole absolute (Kelvin) scale.

Recall Body temp

$37 ° C$ in Fahrenheit? $98.6 ° F$ ::: from $\frac{9}{5} (37) + 32 = 66.6 + 32$ .

See also: Zeroth Law of Thermodynamics, Heat and Internal Energy, Thermal Expansion (mercury/gas thermometers rely on it), and Ideal Gas Law for the $P \propto T$ foundation of the gas thermometer.