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Converting between Celsius and Fahrenheit

A forecast abroad says 25 degrees and a recipe says 425 degrees. How do you turn a reading on one temperature scale into the same temperature on the other?

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Evaluating expressions and formulasMath 6Temperature scales: Celsius, Fahrenheit and KelvinMath 6
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Reading a thermometerTemperature scales: Celsius, Fahrenheit and KelvinTemperatures below zero and temperature change
Converting between Celsius and Fahrenheit

When you reach for it

Most countries give temperatures in degrees Celsius; a few, the United States among them, use degrees Fahrenheit. A weather report in a city you are traveling to, an oven temperature in a recipe from another country, a reading on a medical thermometer, the setting in a freezer manual: each can arrive in one scale while you think in the other.

The two scales measure the same thing but disagree in two ways, both laid out in the lesson on temperature scales. Their zeros are in different places: water freezes at 0^\circ\text{C}, which is 32^\circ\text{F}. And their degrees are different sizes: from freezing to boiling there are 100 Celsius degrees but 212 - 32 = 180 Fahrenheit degrees. So converting is not a single multiplication, the way changing meters to centimeters is. It needs a stretch, for the size of the degree, and a shift, for the place of the zero. The rule that does both is a formula, and using it is substituting the reading you have.

The procedure

Building the rule from two anchors

−40 −20 0 20 60 80 100 212 water boils 98.6 body temperature 37 32 water freezes −40 the scales agree
{}^\circ\mathrm{C}
{}^\circ\mathrm{F}
Figure 1The same temperatures on both scales. Freezing to boiling is 100 Celsius degrees but 180 Fahrenheit degrees, so a Celsius degree is the bigger one, and the zeros sit at different heights.

The freezing and boiling points of water are the two anchors. Between them lie 100 Celsius degrees and 180 Fahrenheit degrees, the same stretch of temperature counted in two sizes of step. So one Celsius degree is worth

\frac{180}{100} = \frac{9}{5} \text{ Fahrenheit degrees,}

which is 1.8 of them. A Celsius degree is the bigger step, which is why the Fahrenheit ticks in the figure are packed more tightly.

Now take a reading of C degrees Celsius. It sits C Celsius degrees above the freezing point (below it when C is negative). Counted in Fahrenheit degrees, that same distance is \displaystyle \frac{9}{5}C: this is the stretch. But the freezing point is not the Fahrenheit zero; it is the Fahrenheit reading 32. So the Fahrenheit reading is \displaystyle \frac{9}{5}C on from 32: this is the shift.

F = \frac{9}{5}C + 32

Both anchors check: C = 0 gives F = 32, and C = 100 gives F = 180 + 32 = 212. When C is negative, \displaystyle \frac{9}{5}C is negative too, and the rule still says the right thing: the reading lies that many Fahrenheit degrees below 32.

Going the other way

A Fahrenheit reading F is known and C is wanted. Then \displaystyle F = \frac{9}{5}C + 32 is an equation in C, and a two-step equation at that. The rule did two things to C: it multiplied by \displaystyle \frac{9}{5}, then added 32. Undo them in the reverse order, doing the same to both sides each time so the two sides stay equal (the properties of equality). Subtract 32 from both sides, then multiply both sides by \displaystyle \frac{5}{9}, the reciprocal of \displaystyle \frac{9}{5}:

\begin{aligned} F &= \tfrac{9}{5}C + 32 \\ F - 32 &= \tfrac{9}{5}C \\ \tfrac{5}{9}(F - 32) &= C \end{aligned}

Turned round, that is the rule for going from Fahrenheit to Celsius:

C = \frac{5}{9}(F - 32)

The brackets carry the order. The 32 was added last, so it comes off first, while the reading is still in Fahrenheit degrees; only then does \displaystyle \frac{5}{9} turn the Fahrenheit degrees above freezing into Celsius degrees. At the boiling point, \displaystyle \frac{5}{9}(212 - 32) = \frac{5}{9}(180) = 100, as it should be.

You can use this second formula, or put the Fahrenheit number into the first one and solve the equation each time. They are the same two steps; Example 2 does it both ways.

Changes of temperature

A rise or a fall is a distance along the scale, not a point on it, and only the size of the degree matters for a distance. Take two Celsius readings C_1 and C_2. Their Fahrenheit readings differ by

\begin{aligned} F_2 - F_1 &= \left(\tfrac{9}{5}C_2 + 32\right) \\ &\quad - \left(\tfrac{9}{5}C_1 + 32\right) \\ &= \tfrac{9}{5}(C_2 - C_1) \end{aligned}

The two 32s cancel. So a change of d Celsius degrees is a change of \displaystyle \frac{9}{5}d Fahrenheit degrees, and a change of e Fahrenheit degrees is \displaystyle \frac{5}{9}e Celsius degrees, with no 32 in either direction. This lesson says “degrees Celsius” for a reading and “Celsius degrees” for a change, as Temperatures below zero and temperature change does.

Worked examples

Example 1

A forecast in Celsius

Nerys is traveling to a city whose weather report gives tomorrow’s high as 25^\circ\text{C}. What is that in degrees Fahrenheit?

Solution

The reading is in Celsius and Fahrenheit is wanted, so substitute C = 25 into the rule that produces F:

F = \frac{9}{5}(25) + 32

Multiply before adding, as the order of operations says. Since 25 \div 5 = 5, \displaystyle \frac{9}{5}(25) = 9 \times 5 = 45. That is the stretch: 25 Celsius degrees above freezing is 45 Fahrenheit degrees above freezing. Then the shift: 45 + 32 = 77. The high is 77^\circ\text{F}, a warm day.

Check by going back: 77 - 32 = 45 and \displaystyle \frac{5}{9}(45) = 25.

Example 2

An oven temperature, two ways

Pascoe’s recipe, from the United States, says to bake at 425^\circ\text{F}. His oven dial is in degrees Celsius, marked every 10 degrees. Which mark should he use?

Solution

First way, with the formula for C:

C = \frac{5}{9}(425 - 32) = \frac{5}{9}(393) = \frac{1965}{9}

The bracket comes first: taking off 32 removes the shift and leaves 393 Fahrenheit degrees above freezing, and \displaystyle \frac{5}{9} turns those into Celsius degrees. Dividing, \displaystyle \frac{1965}{9} = 218.33\ldots, about 218.3^\circ\text{C}.

Second way: put F = 425 into \displaystyle F = \frac{9}{5}C + 32 and solve for C.

425 = \frac{9}{5}C + 32

Subtracting 32 from both sides gives \displaystyle 393 = \frac{9}{5}C. Multiplying both sides by \displaystyle \frac{5}{9} gives \displaystyle C = \frac{5}{9}(393), the same number. The formula for C is nothing more than this solving, done once with a letter instead of a number.

The reading 218.3 lies between the marks 210 and 220, and nearer 220. Pascoe sets the dial to 220^\circ\text{C}. The rounding came at the very end; the working stayed exact until then.

Example 3

Body temperature

About 37^\circ\text{C} is the figure usually given for body temperature. (a) Write it in degrees Fahrenheit. (b) Orsolya’s thermometer reads 102.2^\circ\text{F}. Write this in degrees Celsius, and say how many Celsius degrees above 37^\circ\text{C} it is.

Solution

(a) Substitute C = 37: \displaystyle \frac{9}{5}(37) = \frac{333}{5} = 66.6, and 66.6 + 32 = 98.6. So 37^\circ\text{C} is 98.6^\circ\text{F}.

(b) Now F is known, so use \displaystyle C = \frac{5}{9}(F - 32). The bracket gives 102.2 - 32 = 70.2. Then \displaystyle \frac{5}{9}(70.2) = \frac{351}{9} = 39. The reading is 39^\circ\text{C}, which is 39 - 37 = 2 Celsius degrees above 37^\circ\text{C}. That difference is a change, and it can be read in either scale: 102.2 - 98.6 = 3.6 Fahrenheit degrees, and \displaystyle \frac{5}{9}(3.6) = 2 agrees.

Example 4

A freezer below zero

Marcellus’s freezer manual says to set it to -25^\circ\text{C}. His freezer’s display shows degrees Fahrenheit. What should it read?

Solution

The rule does not change for a reading below zero; the arithmetic just runs through negative numbers (Temperatures below zero and temperature change covers reading them). Substitute C = -25:

F = \frac{9}{5}(-25) + 32

A positive number times a negative one is negative, so \displaystyle \frac{9}{5}(-25) = -45: the freezer is 45 Fahrenheit degrees below the freezing point of water. Adding the shift, -45 + 32 = -13. The display should read -13^\circ\text{F}, below the Fahrenheit zero as well as the Celsius one.

Check: -13 - 32 = -45, and \displaystyle \frac{5}{9}(-45) = -25.

Example 5

A greenhouse warming up

Quintana’s greenhouse is at 12^\circ\text{C} at dawn, and the sun warms it to 27^\circ\text{C} by noon. How many Fahrenheit degrees did it warm?

Solution

The question asks for a change, not a reading. The rise is 27 - 12 = 15 Celsius degrees, and a change uses only the stretch:

\frac{9}{5}(15) = 27

It warmed by 27 Fahrenheit degrees. Converting both readings shows why there is no 32: 12^\circ\text{C} is 21.6 + 32 = 53.6^\circ\text{F} and 27^\circ\text{C} is 48.6 + 32 = 80.6^\circ\text{F}, and 80.6 - 53.6 = 27. The 32 was added to both readings, so it dropped out of their difference.

Example 6

The temperature where the scales agree

Is there a temperature that has the same number on both scales?

Solution

Compare the two numbers at a few readings and watch the gap between them. At the freezing point they are 0 and 32: the Fahrenheit number is ahead by 32. Going warmer will not close that gap. Each Celsius degree warmer adds \displaystyle \frac{9}{5} to the Fahrenheit number but only 1 to the Celsius number, so the Fahrenheit number pulls further ahead, by \displaystyle \frac{9}{5} - 1 = \frac{4}{5} a degree. So go colder, where the gap shrinks by the same \displaystyle \frac{4}{5} a degree. Ten Celsius degrees colder takes 10 off the Celsius number and \displaystyle \frac{9}{5}(10) = 18 off the Fahrenheit number, so the gap closes by 18 - 10 = 8:

C\displaystyle F = \frac{9}{5}C + 32gap F - C
03232
-101424
-20-416
-30-228
-40-400

A gap of 32 closing 8 at a time is gone after 32 \div 8 = 4 steps of 10 degrees, at C = -40. Check: \displaystyle \frac{9}{5}(-40) + 32 = -72 + 32 = -40. So -40^\circ\text{C} and -40^\circ\text{F} are the same temperature, the bottom line in the first figure.

-40
-40
32
(100,\ 212)
F = \tfrac{9}{5}C + 32
F = C
C
F
Figure 2The rule is a straight line. The dashed line F = C meets it once, at (-40, -40).

It is the only one. On the graph the rule is a straight line and “same number” is the dashed line F = C; two straight lines of different steepness cross exactly once. Above -40 the Fahrenheit number is the larger (at freezing, 32 against 0), and below it the smaller: -50^\circ\text{C} is -90 + 32 = -58^\circ\text{F}.

Later, algebra asks the same question as one equation, \displaystyle x = \frac{9}{5}x + 32, with the unknown on both sides; collecting the unknown on one side solves it and gives -40 again.

Boundaries

  • The 32 belongs to readings only. The rules \displaystyle F = \frac{9}{5}C + 32 and \displaystyle C = \frac{5}{9}(F - 32) convert a point on the scale. A change, a rise or a fall, or the gap between two readings converts with \displaystyle \frac{9}{5} or \displaystyle \frac{5}{9} alone.
  • The numbers in the rule are exact. The \displaystyle \frac{9}{5} and the 32 are fixed by how the two scales are defined, not measured, so a conversion is exact until you choose to round. Round only at the end: in Example 2, rounding \displaystyle \frac{5}{9} to 0.56 first would give 0.56 \times 393 = 220.08 rather than 218.3.
  • Converting does not make ratios meaningful. Neither scale starts at “no heat at all”, so one reading being twice another says nothing in either scale; Percent error and Temperatures below zero and temperature change work through what this does to comparisons.
  • There is a floor. No temperature is colder than absolute zero, -273.15^\circ\text{C}, which is -459.67^\circ\text{F}. The formulas accept smaller numbers, but those numbers are not temperatures. The kelvin scale, in Temperature scales: Celsius, Fahrenheit and Kelvin, starts there.

Where this goes next

The rule \displaystyle F = \frac{9}{5}C + 32 is a linear function, and its graph is the straight line in the second figure. The stretch \displaystyle \frac{9}{5} is its rate of change, Fahrenheit degrees per Celsius degree, and the shift 32 is its starting value at C = 0; reading those two numbers off any such rule is Slope-intercept form. A change of temperature, by contrast, is proportional: \displaystyle \frac{9}{5}d has no added number, so its graph runs through the origin, while the readings’ graph misses it by 32. Turning \displaystyle F = \frac{9}{5}C + 32 round into \displaystyle C = \frac{5}{9}(F - 32) is the first case of solving a formula for another letter, which works the same way for area, speed and interest formulas. And the kelvin scale keeps the Celsius degree and moves only the zero, so converting Celsius to kelvin is a shift with no stretch at all.

Exercises

Celsius to Fahrenheit

Convert each reading to degrees Fahrenheit.

  1. The weather report for the city Marcellus is visiting gives a high of 35^\circ\text{C}.
  2. Bath water at 40^\circ\text{C}.
  3. An oven set to 220^\circ\text{C}.
  4. A fever of 38.5^\circ\text{C}.
  5. A winter morning at -16^\circ\text{C}.
  6. A night in a mountain town at -35^\circ\text{C}.

Fahrenheit to Celsius

Convert each reading to degrees Celsius.

  1. A hot-water heater set to 122^\circ\text{F}.
  2. A recipe asking for an oven at 350^\circ\text{F}. Give the answer to one decimal place.
  3. A cold morning at 14^\circ\text{F}.
  4. A thermometer reading of 99.5^\circ\text{F}.
  5. A freezer display reading -0.4^\circ\text{F}.
  6. A desert afternoon at 113^\circ\text{F}.

Changes in temperature

  1. A forecast says tomorrow will be 12 Celsius degrees warmer than today. How many Fahrenheit degrees warmer is that?
  2. An oven heats up by 90 Fahrenheit degrees. By how many Celsius degrees did it heat up?
  3. Over one evening the temperature falls from 13^\circ\text{C} to -22^\circ\text{C}. Find the fall in Fahrenheit degrees, then check it by converting both readings.
  4. Radomir’s temperature falls from 102.4^\circ\text{F} to 99.7^\circ\text{F}. By how many Celsius degrees did it fall?

Comparing and solving

Example 6 found a temperature by watching a gap between two numbers change by a fixed amount for each degree. The same idea answers Exercises 18 to 20.

  1. One city reports 28^\circ\text{C} and another 82^\circ\text{F}. Which is warmer, and by how many Fahrenheit degrees?
  2. At what temperature is the Fahrenheit number exactly twice the Celsius number? Give the reading in both scales.
  3. At what temperature is the Fahrenheit number 50 more than the Celsius number? Give the reading in both scales.
  4. A quick rule for Celsius to Fahrenheit says “double it and add 30”. (a) What does the quick rule give for 25^\circ\text{C}, and how far is that from the exact answer? (b) At what Celsius reading is the quick rule 4 degrees too high?

Explaining why

  1. Two rules are proposed for Fahrenheit to Celsius: \displaystyle C = \frac{5}{9}(F - 32) and \displaystyle C = \frac{5}{9}F - 32. Test both at the boiling point of water, 212^\circ\text{F}. Which is right, and why does the order of the two steps matter?
  2. Show that two Celsius readings C_1 and C_2 have Fahrenheit readings that differ by \displaystyle \frac{9}{5}(C_2 - C_1), and explain in words why converting a change never uses the 32.
  3. Show that converting a Celsius reading C to Fahrenheit and then converting the result back to Celsius always gives C again. Explain which step undoes which.
  4. Explain why there is exactly one temperature at which the two scales show the same number, and why the Fahrenheit number is the larger above it and the smaller below it.
Answers
  1. 95^\circ\text{F}. \displaystyle \frac{9}{5}(35) = 63, and 63 + 32 = 95.
  2. 104^\circ\text{F}. \displaystyle \frac{9}{5}(40) = 72, and 72 + 32 = 104.
  3. 428^\circ\text{F}. \displaystyle \frac{9}{5}(220) = 396, and 396 + 32 = 428.
  4. 101.3^\circ\text{F}. \displaystyle \frac{9}{5}(38.5) = 69.3, and 69.3 + 32 = 101.3.
  5. 3.2^\circ\text{F}. \displaystyle \frac{9}{5}(-16) = -28.8, and -28.8 + 32 = 3.2.
  6. -31^\circ\text{F}. \displaystyle \frac{9}{5}(-35) = -63, and -63 + 32 = -31.
  7. 50^\circ\text{C}. 122 - 32 = 90, and \displaystyle \frac{5}{9}(90) = 50.
  8. About 176.7^\circ\text{C}. 350 - 32 = 318, and \displaystyle \frac{5}{9}(318) = \frac{1590}{9} = 176.66\ldots, which rounds to 176.7.
  9. -10^\circ\text{C}. 14 - 32 = -18, and \displaystyle \frac{5}{9}(-18) = -10.
  10. 37.5^\circ\text{C}. 99.5 - 32 = 67.5, and \displaystyle \frac{5}{9}(67.5) = 37.5.
  11. -18^\circ\text{C}. -0.4 - 32 = -32.4, and \displaystyle \frac{5}{9}(-32.4) = -18.
  12. 45^\circ\text{C}. 113 - 32 = 81, and \displaystyle \frac{5}{9}(81) = 45.
  13. 21.6 Fahrenheit degrees. A change uses only the stretch: \displaystyle \frac{9}{5}(12) = 21.6.
  14. 50 Celsius degrees. A change uses only the stretch: \displaystyle \frac{5}{9}(90) = 50.
  15. A fall of 63 Fahrenheit degrees. The fall is 13 - (-22) = 35 Celsius degrees, and \displaystyle \frac{9}{5}(35) = 63. Check: 13^\circ\text{C} is 23.4 + 32 = 55.4^\circ\text{F} and -22^\circ\text{C} is -39.6 + 32 = -7.6^\circ\text{F}, and 55.4 - (-7.6) = 63.
  16. 1.5 Celsius degrees. The fall is 102.4 - 99.7 = 2.7 Fahrenheit degrees, and \displaystyle \frac{5}{9}(2.7) = 1.5.
  17. The 28^\circ\text{C} city, by 0.4 Fahrenheit degrees. \displaystyle \frac{9}{5}(28) = 50.4 and 50.4 + 32 = 82.4, so 28^\circ\text{C} is 82.4^\circ\text{F}, and 82.4 - 82 = 0.4.
  18. 160^\circ\text{C}, which is 320^\circ\text{F}. At 0^\circ\text{C} the Fahrenheit number is 32 and twice the Celsius number is 0, so the Fahrenheit number is ahead by 32. Each Celsius degree warmer adds \displaystyle \frac{9}{5} to the Fahrenheit number and 2 to twice the Celsius number, so the gap closes by \displaystyle 2 - \frac{9}{5} = \frac{1}{5}. Closing 32 takes \displaystyle 32 \div \frac{1}{5} = 160 degrees. Check: \displaystyle \frac{9}{5}(160) + 32 = 288 + 32 = 320, which is 2(160).
  19. 22.5^\circ\text{C}, which is 72.5^\circ\text{F}. At 0^\circ\text{C} the gap F - C is 32, and each Celsius degree warmer adds \displaystyle \frac{4}{5} to it, as in Example 6. It has to grow by 50 - 32 = 18, which takes \displaystyle 18 \div \frac{4}{5} = 18 \times \frac{5}{4} = 22.5 degrees. Check: \displaystyle \frac{9}{5}(22.5) + 32 = 40.5 + 32 = 72.5, which is 22.5 + 50.
  20. (a) The quick rule gives 2(25) + 30 = 80, while the exact rule gives \displaystyle \frac{9}{5}(25) + 32 = 77, so the quick rule is 3 degrees too high. (b) 30^\circ\text{C}. At 0^\circ\text{C} the quick rule gives 30 against the exact 32, so it is 2 too low. Each Celsius degree warmer, the quick rule adds 2 and the exact rule \displaystyle \frac{9}{5}, so the quick rule gains \displaystyle \frac{1}{5}. Going from 2 too low to 4 too high is a gain of 6, which takes \displaystyle 6 \div \frac{1}{5} = 30 degrees. Check: the quick rule gives 2(30) + 30 = 90 and the exact rule \displaystyle \frac{9}{5}(30) + 32 = 86, and 90 - 86 = 4.
  21. The first. It gives \displaystyle \frac{5}{9}(212 - 32) = \frac{5}{9}(180) = 100, the boiling point. The second gives \displaystyle \frac{5}{9}(212) - 32 = \frac{772}{9} \approx 85.8, which is wrong. The rule for Fahrenheit multiplies by \displaystyle \frac{9}{5} and then adds 32, so undoing it takes off the 32 first. The 32 is a count of Fahrenheit degrees; once the number has been multiplied by \displaystyle \frac{5}{9} it is in Celsius degrees, and taking 32 of those away removes the wrong amount.
  22. F_2 - F_1 is \displaystyle \left(\frac{9}{5}C_2 + 32\right) - \left(\frac{9}{5}C_1 + 32\right), which is \displaystyle \frac{9}{5}(C_2 - C_1), since the two 32s cancel and \displaystyle \frac{9}{5} factors out by the distributive property. In words: the shift moves every reading up by the same 32, so it moves both ends of a change together and leaves the distance between them alone. Only the size of the degree affects a distance.
  23. Back in Celsius, the reading is \displaystyle \frac{5}{9}\left(\left(\frac{9}{5}C + 32\right) - 32\right), and this is \displaystyle \frac{5}{9} \cdot \frac{9}{5}C = C. Subtracting 32 undoes adding 32, and multiplying by \displaystyle \frac{5}{9} undoes multiplying by \displaystyle \frac{9}{5}, because \displaystyle \frac{5}{9} \cdot \frac{9}{5} = 1. The steps are undone in the reverse order to the one they were done in. Starting from Fahrenheit works the same way: \displaystyle \frac{9}{5} \cdot \frac{5}{9}(F - 32) + 32 = F.
  24. The gap between the two numbers is \displaystyle F - C = \frac{9}{5}C + 32 - C = \frac{4}{5}C + 32. Each Celsius degree warmer adds \displaystyle \frac{4}{5} to this gap, so the gap only ever grows as the temperature rises, and it can be zero at one temperature only: \displaystyle \frac{4}{5}C + 32 = 0 at C = -40. Above -40 the gap is positive, so the Fahrenheit number is larger; below it the gap is negative and the Fahrenheit number is smaller. On the graph, the rule and the line F = C are straight lines of different steepness, and such lines cross exactly once.
kg1128 · updated 2026-10-07 · CC-BY-SA-4.0