Tough conversion of temperature (C to F) question?

Simple maths. F-32 divided by 9 = C divided by 5. I never was much good at advanced maths, but that one is easy, and these days you have a calculator, which I did not have in the 1960s.

A Centigrade degree is 1.8 times the size of a Fahrenheit degree. When you change temperature by 1 degree C, you change temperature by 1.8 degrees F. When you change temperature by 10 degrees C, you change temperature by 10*1.8 = 18 degrees F.

When you change from 160 C to 180 C, that's a change of 20 C which is 20 * 1.8 = 36 Fahrenheit degrees. Which is the difference between 356 F and 320 F, exactly as it should be.

Why is that bothering you? What are you expecting?

Look, let's start at the freezing point of water, which is 32 F and 0 C. Let's go up to 160 C. That's a change of 160 * 1.8 = 288 Fahrenheit degrees, so we're now at 32 + 288 = 320 F. Is that a different number than you were expecting? Why?

And now go another 20 degrees C, which is 20*1.8 = 36 degrees F, so now we're at 160 + 20 = 180 C and 320 + 36 = 356 F.

They don't have the same zero, you realize that, right? When you're at 0 C you're not at 0 F. When you're at 0 F you're not at 0 C. But from any starting point, such as 0 C / 32 F or 100 C / 212 F, or -40 C / -40 F (yeah, that's the only temperature where they're the same number), every step of one celsius degree is 1.8 steps of fahrenheit degrees.

The issue why it's difficult is that both scales DO NOT intersect at 0.

They are linear, they just intersect at -40 C&F degrees.

f = (9/5)c + 32, which for 160 C is 320 F. Somebody said it's not linear. Of course it is! It's the epitome of a linear equation...

y = mx + b, the slope 9/5, the intercept 32.

Answer: You just got the conversion wrong. Not a big deal. I do that all the time.

Positions on the Celsius or Fahrenheit scales

and

Distances on the Celsius or Fahrenheit scales

are not the same thing.

Boston is 200 miles from New York City.

Hartford is 100 miles from New York.

Does Boston minus Hartford equal Hartford?

And what if we measure from Los Angeles? Would Boston minus Hartford still equal Hartford then?

(for simplicity, assume these cities are all in a straight line.)

The Celsius-Fahrenheit conversion that you know is equivalent to converting "miles from New York" to "km from Los Angeles." The two scales not only use units of different size, they also measure from different "zero" locations.

The distance between Hartford and Boston is 100 miles.

The distance between Hartford and Boston is 160 km, because 100 miles is the same as 160 km.

So Hartford is also 160 km from New York; but neither scale is measuring km from New York! Hartford is not 160 km from Los Angeles.

The distance between 160°C and 180°C is the same as the distance between 0°C and 20°C.

Or in Fahrenheit: the distance between 320°F and 356°F is the same as the distance between 32°F and 68°F.

But there's no reason that distance should be 68°F.

You do realize that the heat being measured is exactly the same, right? That the only difference is the index reading?

A football field, whether measured out as 100 yards or 91.44 meters is still the same length.

because the temperature conversion is not linear.

it't not ºC = kºF, it's ºC = kºF + b

Here is a simpler example.

(boiling and freezing points)

212ºF + 32ºF = 244ºF

converting each to ºC

100ºC + 0ºC = 100ºC

but 100ºC does not equal 244ºF

or in kelvins

373K + 273K = 646K

but 646K does not equal either 344ºF or 100ºC

Because 358 F minus 68 F equals 290 F.

This is a simple case of subtraction: Does 358 - 68 = 320?

One of the fundamental things that one has to learn in science is how to weed out all of the information that is irrelevant to the question. Since your question only concerns temperatures in the Fahrenheit scale, any conversion crap is totally unrelated to the question at hand.

Don't confuse mathematical functions with ratios.