Celsius vs Kelvin: What Is the Difference?

The distinction, stated once

One number separates the two scales, and one property is shared.

The degree is the same size. Warming something by one kelvin and warming it by one degree Celsius are the same physical change. So a difference of temperature carries the same number on both scales: a rise of $60\ \text{K}$ is a rise of $60$ degrees Celsius, exactly, with no conversion.

The zeros are different. The kelvin scale starts at absolute zero. The Celsius scale starts about $273$ kelvin higher, at the freezing point of water. The conversion is

usually rounded to $273$ in physics work. That conversion is not printed on the AP Physics 2 equation sheet and the CED does not state it, so it is worth knowing rather than looking up: the sheet's unit-symbol table lists both degree Celsius and kelvin as units and leaves the relationship between them to you.

Everything that follows is a consequence of those two facts together. Because the degree is shared, a $\Delta T$ is scale-free. Because the zero is not, a $T$ is not. So the question to ask of any equation is not "which scale should I use" but "does this equation contain $T$ or $\Delta T$?" Answer that and the scale question answers itself.

The reason kelvin exists at all is that some quantities are genuinely proportional to temperature, and a proportionality needs a scale whose zero means zero. 9.1.B.1 makes temperature the average kinetic energy of the atoms in a system, and 9.1.B.1.ii writes that as $K_{\text{avg}} = \frac{3}{2}k_B T = \frac{1}{2}mv_{\text{rms}}^2$. Halve $T$ on the kelvin scale and you halve the average kinetic energy. Halve a Celsius reading and you have done arithmetic on an arbitrary offset.

Celsius against kelvin, row by row

Row eight is the tell that saves the most marks, and it needs no memorising. Look at the constants the AP Physics 2 sheet prints: the universal gas constant is $8.31\ \text{J/(mol} \cdot \text{K)}$, Boltzmann's constant is $1.38 \times 10^{-23}\ \text{J/K}$, Wien's constant is $2.90 \times 10^{-3}\ \text{m} \cdot \text{K}$ and the Stefan-Boltzmann constant is $5.67 \times 10^{-8}\ \text{W}/(\text{m}^2 \cdot \text{K}^4)$. Every one of them is quoted per kelvin, or per kelvin to a power. A constant defined that way only reproduces measured behaviour when the temperature you feed it is measured from absolute zero.

Row four is worth pausing on because it is the fastest error check available. If a calculation hands you a negative absolute temperature, you have either substituted a Celsius reading or made a sign slip. There is no third option.

And row nine is a real convention, not a stylistic preference: the unit is the kelvin and it takes no degree sign, so $300\ \text{K}$ rather than $300$ degrees K.

Which printed equations need kelvin, checked one at a time

The AP Physics 2 equation sheet's Thermal Physics panel prints eight equations. Here they are, sorted by what they do with temperature.

Three take an absolute temperature and require kelvin:

1. $K_{\text{avg}} = \frac{3}{2}k_B T = \frac{1}{2}mv_{\text{rms}}^2$, from 9.1.B.1.ii.
2. $PV = nRT = Nk_B T$, from 9.2.A.2.
3. $U = \frac{3}{2}nRT = \frac{3}{2}Nk_B T$, from 9.4.A.1.ii.

Two take a temperature difference and work in either scale:

1. $Q = mc\Delta T$, from 9.5.A.1.
2. $\dfrac{Q}{\Delta t} = \dfrac{kA\Delta T}{L}$, from 9.5.B.1.

Three contain no temperature at all:

1. $P = \dfrac{F_{\perp}}{A}$, from 9.1.A.1.ii.
2. $W = -P\Delta V$, from 9.4.B.1.iii.
3. $\Delta U = Q + W$, from 9.4.B.1.ii.

That is the whole panel, three plus two plus three.

Two more printed equations sit outside Unit 9 and belong on the absolute list. The Modern Physics panel prints Wien's displacement relation $\lambda_{\text{max}} = b/T$ and the Stefan-Boltzmann radiated power $P = A\sigma T^4$, and the symbol list on that panel does not say "temperature": it says $T$ is the absolute temperature. The Thermal Physics panel's own symbol list simply says temperature, which is the one place the sheet leaves the requirement implicit rather than stating it.

A warning about the letter $T$ while you are reading the sheet. In the Thermal Physics panel it is a temperature. In the Modern Physics panel it is an absolute temperature. In the Waves, Sound, and Optics panel it is a period, in seconds, which is why $T = 1/f$ appears there. Three panels, three meanings, one letter. Read the panel.

The case that separates them: two temperatures, two calculations, one disaster

Take a sample of gas and warm it from $20$ degrees Celsius to $80$ degrees Celsius. Two questions, both about the same two temperatures.

Question one: how much energy does it take to warm $2.0\ \text{kg}$ of water through the same range? Use $Q = mc\Delta T$ with $c = 4186\ \text{J/(kg} \cdot \text{K)}$.

• In Celsius: $\Delta T = 80 - 20 = 60$ degrees Celsius, so $Q = (2.0)(4186)(60) = 5.02 \times 10^5\ \text{J}$.
• In kelvin: $\Delta T = 353 - 293 = 60\ \text{K}$, so $Q = 5.02 \times 10^5\ \text{J}$.

Identical, to every digit, because the equation contains a difference and the two scales share a degree size.

Question two: by what factor does the root-mean-square speed of the gas rise over the same range? From 9.1.B.1.ii, $\frac{1}{2}mv_{\text{rms}}^2 = \frac{3}{2}k_B T$, so $v_{\text{rms}} \propto \sqrt{T}$ with $T$ absolute.

• Correctly, in kelvin: $\sqrt{353/293} = \sqrt{1.2048} = 1.098$, a rise of about $9.8$ percent.
• Incorrectly, in Celsius: $\sqrt{80/20} = \sqrt{4} = 2.000$, a doubling.

The correct answer and the Celsius answer differ by a factor of $1.82$. Nothing changed between the two calculations except the scale the numbers were read on, and one of the two equations did not care while the other was destroyed.

That is the distinction with a consequence, and it explains why the rule is not "always use kelvin". Both answers to question one are right. The rule that actually works is: look for a $\Delta$. If the temperature appears only inside a difference, either scale is fine and Celsius is often more convenient. If it appears on its own, convert first, every time.

One more asymmetry hides in that. The Celsius error in question two is not a fixed offset you could correct at the end; it is a different ratio, and how wrong it is depends on where you are on the scale. The same mistake between $200$ and $260$ degrees Celsius would give $\sqrt{260/200} = 1.140$ against the correct $\sqrt{533/473} = 1.062$, an error of $7$ percent rather than $82$. There is no way to spot the mistake from the size of the answer.

Why absolute zero is where it is

The kelvin scale's zero is not a convention. AP Physics 2 gives it an operational definition you can carry out with a pressure gauge, and it is one of the more elegant things in Unit 9.

Essential knowledge 9.2.A.4 says a temperature at which an ideal gas has zero pressure can be extrapolated from a graph of pressure as a function of temperature. 9.2.A.3 sets it up, saying graphs modeling the pressure, temperature, and volume of gases can be used to describe or determine properties of that gas.

Here is the procedure. Seal a fixed volume of gas, measure its pressure at several temperatures, and plot pressure against temperature in degrees Celsius. The points fall on a straight line, but the line does not pass through the origin: at $0$ degrees Celsius the gas still has plenty of pressure. Extend the line backwards to where the pressure would be zero, and it crosses the axis at about $-273$ degrees Celsius.

Do the same experiment with a different gas, a different volume or a different amount, and the slope changes but the intercept does not. That intercept is absolute zero, and the kelvin scale is defined as the same degree size with its origin moved there. Which is why $PV = nRT$ works in kelvin: on that scale the line does pass through the origin, so pressure really is proportional to temperature, and 9.2.A.2's equation is a proportionality rather than a straight line with an offset.

The kinetic picture from 9.1 fits the same point. 9.1.B.1 makes temperature the average kinetic energy of the atoms and 9.1.B.1.ii ties it to $v_{\text{rms}}$, so zero temperature on the kelvin scale corresponds to zero average translational kinetic energy, which is also zero pressure, since 9.1.A.1 explains pressure as arising from atoms colliding with the container.

One restraint on the argument. All of this is stated for an ideal gas, and 9.2.A.1 lists what that model assumes: instantaneous velocities of atoms are random, the volume of the atoms is negligible compared to the total volume occupied by the gas, the atoms collide elastically, and the only appreciable forces on the atoms are those that occur during collisions. Real gases liquefy long before the extrapolation gets there. The line is the evidence for where absolute zero is; it is not a claim that you could take a gas to it.

When it costs a mark

• Substituting a Celsius reading into $PV = nRT$. This is the whole family of errors in one, and it is not a small numerical slip: at room temperature it is off by a factor of about $15$, and near freezing it divides by something close to zero.
• Converting a temperature difference. Adding $273$ to a $\Delta T$ is the mirror-image mistake. A rise of $60$ degrees Celsius is a rise of $60\ \text{K}$, not $333\ \text{K}$. Both $Q = mc\Delta T$ and the conduction rate equation take differences.
• Converting the specific heat. Because the degree size is shared, $4186\ \text{J/(kg} \cdot \text{K)}$ and $4186\ \text{J/(kg} \cdot ^{\circ}\text{C)}$ are the same number, and thermal conductivities work the same way. There is nothing to convert.
• Halving or doubling a Celsius temperature in a proportional-reasoning question. "The temperature doubles" means the kelvin temperature doubles. From $20$ degrees Celsius that is $293\ \text{K}$ to $586\ \text{K}$, which is $313$ degrees Celsius, not $40$.
• Writing kelvin with a degree sign, or calling it degrees Kelvin. The unit is the kelvin, written K.
• Reading $T$ as a temperature in a waves question. In the sheet's Waves, Sound, and Optics panel, $T$ is the period in seconds and $T = 1/f$ is the printed relation. Nothing about that line involves temperature.
• Using $273$ when a question wants more precision. The conversion is $273.15$, and in most AP calculations the difference is well below the significant figures you are working to. It is worth checking rather than assuming: if two temperatures are close together, the rounding can move a ratio in the third digit.
• Reporting a negative kelvin temperature. There is no such thing on this scale. It is a signal that a Celsius value went in somewhere, or that a subtraction ran the wrong way.

When they coincide, and why that lulls you

They coincide in one important place and one misleading one.

The important one: any equation written in $\Delta T$. Two of the eight printed thermal equations are like that, and they cover most of the calorimetry a student ever does. So a whole term of laboratory work can be completed in Celsius, correctly, without the scale ever mattering. That is not a loophole; it is a genuine property of those equations. But it builds a habit that fails the first time an ideal-gas question arrives.

The misleading one: large temperatures. At $1000$ degrees Celsius the kelvin value is $1273\ \text{K}$, a difference of $27$ percent, which is enough to be wrong and not obviously enough to look wrong. At $20$ degrees Celsius the same offset is a factor of nearly $15$, which any sanity check would catch. So the error is loudest exactly where students are least likely to make it and quietest where the numbers are big, which is the wrong way round for a mistake to behave.

There is a third coincidence in how questions are written. Exam questions state temperatures in whichever unit is natural for the situation, so a calorimetry question may be in Celsius and a gas-law question in kelvin, in the same problem. The habit that survives that is to convert every temperature to kelvin as you read it, and then use the Celsius-friendliness of $\Delta T$ only as a check rather than as a shortcut. Converting when there is no need costs nothing; failing to convert when there is costs the question.

The related distinction is between the temperature and the transfer it drives, which is heat vs temperature, and the two processes where temperature is either fixed or free are in isothermal vs adiabatic.