Q.The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Concept understanding — Temperature Scale Conversion
Temperature Scale Conversion
You already know temperature as a measure of hotness or coldness. But here's the thing: different countries and different scientific fields measure that same hotness using different numbers. Water boils at 100 on the Celsius scale, but at 212 on the Fahrenheit scale. That's not because the water is different — it's because the rulers are different.
The core idea
A temperature scale is just a number line someone decided to draw on the same physical reality. Converting between scales means finding the number on one number line that corresponds to the same physical hotness as a given number on another number line.
Think of it like this: if you measure a table's length in feet and get 6, and I measure it in metres and get 1.83, we're both describing the same table. Temperature conversion works exactly the same way — same physical state, different numbers.
The three main scales you'll meet
| Scale | Freezing point of water | Boiling point of water | Used by |
|---|---|---|---|
| Celsius (°C) | 0 | 100 | Most of the world, science |
| Fahrenheit (°F) | 32 | 212 | USA, a few other countries |
| Kelvin (K) | 273.15 | 373.15 | All scientific work |
Notice something: Celsius and Kelvin have the same step size — a change of 1°C is exactly a change of 1 K. They just start at different places. Fahrenheit has a smaller step size (180 steps between freezing and boiling, instead of 100).
The conversion formulas
°F=59(°C)+32
°C=95(°F−32)
K=°C+273.15
These aren't magic. Each one comes from the simple idea of matching two number lines.
Why the formulas look like that
Between freezing and boiling water:
- Celsius has 100 steps (0 to 100)
- Fahrenheit has 180 steps (32 to 212)
So one Celsius step is 100180=59 Fahrenheit steps. That's the 59 factor. The +32 just shifts the starting point — because 0°C doesn't correspond to 0°F, it corresponds to 32°F.
For Kelvin: since the step size is identical to Celsius, you just add the offset. The 273.15 comes from the fact that absolute zero (the coldest possible temperature) is 0 K, which is −273.15°C.
A worked example
Convert 25°C to Fahrenheit.
Step 1: Multiply by 59.
25×59=25×1.8=45
Step 2: Add 32.
45+32=77
So 25°C = 77°F. A warm spring day in Celsius terms is 77°F — same weather, different number.
For quick mental estimates: double the Celsius and add 30. It's not exact (you get 80 instead of 77 here), but it's close enough for everyday "should I wear a jacket?" decisions.
The one thing students mess up
Never just add or subtract without the factor. "It's 30°C outside, so it must be 30 + 32 = 62°F" is wrong. You must multiply by 59 first. The 32 is a shift after scaling, not before.
Why Kelvin matters
Kelvin is the scientist's scale because it starts at absolute zero — the point where particles have minimum possible thermal motion. This makes all gas laws and thermodynamic equations simple. You'll never see a negative Kelvin temperature in normal physics; 0 K is the floor.
When a problem gives you Celsius and the formula uses Kelvin (like the ideal gas law PV=nRT), you must convert: K=°C+273.15.
The big picture
Temperature scale conversion is just relabelling the same physical reality. The formulas are linear — multiply to adjust step size, add to adjust zero point. Once you see that, every conversion is just arithmetic.
For quick revision, remember that Temperature Scale Conversion is drawn directly from the Thermal Properties of Matter coverage of the NCERT/CBSE Class 11 Physics syllabus and recurs often in JEE Main and NEET papers, which is exactly why "Temperature Scale Conversion important questions" shows up so often in Physics question banks. The clearest way to build exam confidence here is to combine this explanation with the NCERT Physics textbook's own solved examples and chapter-end questions.
Use TC=TK−273.15 and TF=59TC+32.
Neon (24.57 K): TC=24.57−273.15=−248.58∘C; TF=59(−248.58)+32=−415.44∘F.
Carbon dioxide (216.55 K): TC=216.55−273.15=−56.60∘C; TF=59(−56.60)+32=−69.88∘F.
Neon: −248.58∘C, −415.44∘F. Carbon dioxide: −56.60∘C, −69.88∘F.
Converting Kelvin to Celsius using TC=TK−273.15, then to Fahrenheit using TF=59TC+32: neon's triple point is −248.58∘C (−415.44∘F), and carbon dioxide's is −56.60∘C (−69.88∘F).
All three temperature scales measure the same physical hotness, just with different zero points and step sizes. Kelvin and Celsius have the same step size (a change of 1 K equals a change of 1∘C), differing only by an offset of 273.15. Fahrenheit has a smaller step size, related to Celsius by TF=59TC+32.
Neon: TK=24.57 K
To Celsius:
TC=24.57−273.15=−248.58 ∘C.
To Fahrenheit:
TF=59(−248.58)+32=−447.44+32=−415.44 ∘F.
This extreme cold matches neon's role as a noble gas with very weak intermolecular attraction - it stays gaseous down to almost absolute zero.
Carbon dioxide: TK=216.55 K
To Celsius:
TC=216.55−273.15=−56.60 ∘C.
To Fahrenheit:
TF=59(−56.60)+32=−101.88+32=−69.88 ∘F.
This is well below room temperature - consistent with CO2 subliming directly from solid to gas ("dry ice") at ordinary atmospheric pressure rather than melting.
| Substance | Kelvin | Celsius | Fahrenheit |
|---|---|---|---|
| Neon | 24.57 K | −248.58∘C | −415.44∘F |
| Carbon dioxide | 216.55 K | −56.60∘C | −69.88∘F |
Use the precise offset 273.15, not the rounded 273 - small in most problems, but it's good practice to keep the full precision here.
Neon: −248.58∘C and −415.44∘F. Carbon dioxide: −56.60∘C and −69.88∘F.
Since both conversions trace back to the same Kelvin value, you can skip the intermediate Celsius step and go from Kelvin straight to Fahrenheit in one formula: TF=59(TK−273.15)+32. For neon, TF=59(24.57−273.15)+32=−415.44∘F; for CO2, TF=59(216.55−273.15)+32=−69.88∘F — same results, one algebra step shorter. The physical gut-check worth remembering: an entire 273.15 K offset is 'used up' just reaching 0∘C, so neon's triple point, barely 25 K above absolute zero, is bound to land at an extreme negative Fahrenheit value — the huge negative numbers are a sign the physics is right, not an arithmetic slip.
Showing the 12 most recent of 22 on this concept.
- CBSE 2026Set ANNUAL1 markQ.Fill in the blank: The boiling point of water is ______ Celsius at 1.013 bar pressure.
›Reveal solutionSolution
Water's normal boiling point, at standard atmospheric pressure of 1.013 bar, is 100°C (373.15 K).
The boiling point of a liquid is the temperature at which its vapour pressure equals the external (atmospheric) pressure. This value depends on the surrounding pressure — at lower pressures (e.g., high altitude) water boils at a lower temperature, and at higher pressures it boils at a higher temperature. The standard atmospheric pressure is defined as 1.013 bar (1 atm, or 101.325 kPa), and at this pressure the boiling point of pure water is exactly 100°C.
✓Final answerThe boiling point of water at 1.013 bar pressure is 100°C.
- CBSE 2026Set ANNUAL1 markMCQQ.The temperature at which Celsius temperature and Fahrenheit temperature are equal, is(a) 0°(b) 40°(c) − 40°(d) − 32°
›Reveal solutionSolution
Celsius and Fahrenheit are equal at -40 degrees. Answer (C).
The conversion is F = (9/5)C + 32. Set F = C = x:
x = (9/5)x + 32
x - (9/5)x = 32
-(4/5)x = 32
x = -40.
So at -40, the Celsius and Fahrenheit readings are the same (-40 C = -40 F).
✓Final answer(C) -40 degrees.
- CBSE 2025Set ANNUAL1 markMCQQ.The temperature which have same numerical value in Celsius and Fahrenheit scales is (A) 40° (B) -40°C (C) 100° (D) 273°
›Reveal solutionSolution
The Celsius and Fahrenheit scales read the same numerical value at -40°.
The relation between Celsius (C) and Fahrenheit (F) scales is:
5C=9F−32
Setting C=F (equal numerical value) and solving:
5C=9C−32⇒9C=5C−160⇒4C=−160⇒C=−40
So at −40°, both scales show the same reading: −40°C=−40°F.
✓Final answer(B) -40°C.
- CBSE 2025Set hz1 markMCQQ.On the basis of absolute zero as the scale of temperature, the ice melts at:(a) 100 degrees(b) 273.15 K(c) 373.15 K(d) None of them
›Reveal solutionSolution
The Kelvin scale is defined so that its zero coincides with absolute zero, and the ice point (0 degree C, where ice melts under standard pressure) is 273.15 K on this scale.
The absolute (Kelvin) temperature scale is related to the Celsius scale by:
T(K) = T(degree C) + 273.15
The melting point of ice (the ice point) under standard atmospheric pressure is 0 degree C by definition of the Celsius scale. Converting this to the Kelvin scale:
T(K) = 0 + 273.15 = 273.15 K
By contrast, 373.15 K corresponds to 100 degree C, the boiling point of water at standard pressure -- the value in option (c) is the steam point, not the ice point.
✓Final answerThe correct option is (b) 273.15 K.
- CBSE 2024Set ANNUAL1 markMCQQ.The temperature of a body is found in Kelvin scale as x K and in Fahrenheit scale x°F. The value of x will be (A) 574.25 (B) 301.25 (C) 335 (D) 365
›Reveal solutionSolution
The temperature at which the Kelvin and Fahrenheit numerical values coincide is x=574.25.
Conversion: °F=59(K−273)+32. Setting both readings equal to x: x=59(x−273)+32.
Multiply by 5: 5x=9(x−273)+160=9x−2457+160=9x−2297.
5x−9x=−2297⇒−4x=−2297⇒x=574.25.
✓Final answer(A) 574.25.
- CBSE 2024Set SET-AP55001 markMCQQ.The relation between the Centigrade (Celsius) temperature scale and the Kelvin temperature scale is:(a) c = k - 273(b) c = k + 273(c) c = k + 100(d) c = k - 100
›Reveal solutionSolution
The Celsius and Kelvin scales have the same size of degree but different zero points: 0°C = 273 K (more precisely 273.15 K), so C = K − 273.
The Kelvin scale starts at absolute zero (0 K), while the Celsius scale sets 0°C at the freezing point of water. The freezing point of water is 273.15 K, so to convert from Kelvin to Celsius you subtract 273 (approximately):
C = K − 273
Check: at K = 373 (boiling point of water), C = 373 − 273 = 100°C — correct. At K = 273, C = 0°C — correct.
✓Final answerThe correct option is (a) c = k − 273.
- CBSE 2024Set SET-AP55001 markQ.At what temperature will the readings of the Fahrenheit and Celsius scales be equal?
›Reveal solutionSolution
The Fahrenheit and Celsius scales give the same numerical reading at exactly −40 degrees.
The conversion between the two scales is:
F = (9/5)C + 32
Setting F = C (both readings equal, call the common value x):
x = (9/5)x + 32
x − (9/5)x = 32
−(4/5)x = 32
x = 32 × (−5/4) = −40
So at −40°, the Celsius and Fahrenheit thermometers show the identical numerical value: −40°C = −40°F.
✓Final answer−40° (−40°C = −40°F).
- CBSE 2023Set ANNUAL1 markMCQQ.SI unit of temperature is(a) Kelvin(b) Celsius(c) Fahrenheit(d) Centigrade
›Reveal solutionSolution
Kelvin is the SI base unit for temperature; Celsius, Fahrenheit and Centigrade are common but non-SI temperature scales.
The International System of Units (SI) defines seven base units, one for each fundamental physical quantity. For temperature, the SI base unit is the kelvin (K), defined via the thermodynamic (absolute) temperature scale, with 0 K = absolute zero. Celsius (°C) is an SI-accepted derived scale (T(K) = T(°C) + 273.15) but is not itself the SI base unit; Fahrenheit is not part of the SI system at all. "Centigrade" is simply an older name for the Celsius scale.
✓Final answer(a) Kelvin.
- CBSE 2023Set ANNUAL1 markMCQQ.Temperature difference of 20°C is equivalent to:(a) 20 K(b) 293 K(c) 253 K(d) 0 K
›Reveal solutionSolution
Celsius and Kelvin scales have identically sized degrees, so a temperature DIFFERENCE of 20 degC is exactly 20 K -- only absolute temperature readings differ by the 273 offset, not temperature intervals.
The relation between an absolute Celsius reading and Kelvin reading is T(K) = T(degC) + 273. For two temperatures T1 and T2 in Celsius, converting both to Kelvin:
T1(K) = T1(degC) + 273
T2(K) = T2(degC) + 273
Subtracting: T2(K) - T1(K) = T2(degC) - T1(degC)
So the DIFFERENCE is unchanged by the conversion -- the +273 offset cancels out. A temperature CHANGE of 20 degC is therefore also a change of 20 K. (293 K and 253 K would be relevant if this were an absolute temperature reading of 20 degC or -20 degC, not a difference.)
✓Final answerThe correct option is (a) 20 K.
- CBSE 2023Set ANNUAL1 markMCQQ.Absolute zero of temperature is:(a) 100°C(b) 273°C(c) −273°C(d) −273.15°C
›Reveal solutionSolution
Absolute zero = −273.15 °C = 0 K.
The Kelvin (absolute) scale is related to Celsius by T(K) = t(°C) + 273.15. At absolute zero T = 0 K, so t = −273.15 °C. This is the lowest temperature theoretically attainable, at which molecular motion is minimum.
✓Final answer(D) −273.15°C.
- CBSE 2023Set ANNUAL1 markMCQQ.Temprature which has equal reading on Celcius and Faherenhit scale:(a) 0°(b) 30°(c) −40°(d) 40°
›Reveal solutionSolution
The two scales agree at −40°.
The Celsius–Fahrenheit relation is F = (9/5)C + 32. For equal readings put F = C = x:
x = (9/5)x + 32 ⇒ x − (9/5)x = 32 ⇒ −(4/5)x = 32 ⇒ x = −40.
So −40 °C = −40 °F.
✓Final answer(C) −40°.
- CBSE 2022Set TERM11 markMCQQ.The temperature of a body in Celsius scale is 40°C. The temperature in Farenheit scale is(1) 32° F(2) 72° F(3) 104° F(4) None of these
›Reveal solutionSolution
The Celsius-to-Fahrenheit conversion formula, F = (9/5)C + 32, directly converts 40°C to 104°F.
The relationship between the Celsius (C) and Fahrenheit (F) temperature scales is:
F = (9/5) C + 32
Substituting C = 40°C:
F = (9/5)(40) + 32 = (9 x 40)/5 + 32 = 360/5 + 32 = 72 + 32 = 104
So 40°C corresponds to 104°F.
✓Final answer(3) 104° F.
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