# Celsius to Fahrenheit Conversion

Consider the following question: "The maximum temperature of a day is 42 degrees celsius. The minimum temperature is 28 degrees celsius. What is the difference of these temperatures on the Fahrenheit scale?"

When I calculate the answer in two different methods, I get two different answers. I'd like to know why.

Method 1: Subtract 28 degrees celsius from 42 degrees celsius. Convert the resulting answer to fahrenheit. This method yields an answer of - 57.6 degrees celsius

Method 2: First convert 42 degrees celsius to fahrenheit. Then convert 28 degrees celsius to fahrenheit. Then find the difference of the resulting two numbers. This method yields an answer of 25.2 degrees fahrenheit.

Can somebody clearly explain what's happening here?

## 4 Answers

Your Method 1 is wrong. They asked for a range not a quantity. Method 1 gives you the Fahrenheit value for $14^o \text{C},$ which is $57.2^o \text{F}$

$$14^o \text{C}=57.2^o \text{F}~~and~~42^o \text{C}=107.6^o \text{F}$$ $$\therefore~~\text{Difference}=50.4^o \text{ F}$$

You can also find the difference by $$F_1 - F_2=\frac{(C_1 - C_2)\times 9}{5}$$ $$F_1 - F_2=\frac{(28)\times 9}{5}=50.4$$

Basically, neother the function that takes Fahrenheit to degrees nor the other way around is a linear function (strictly-speaking, it is not an additive function, meaning $f(a+b) \neq f(a)+f(b)$). Let F be the function that takes temperature in Celsius to Fahrenheit. You just described a case of how $F(a-b)\neq F(a)-F(b)$. Actually, both F, and its inverse $F^{-1}$ are affine maps, but not linear/additive ones; Affine functions are not additive.

Since you have for the conversion from Celsius to Fahrenheit $$F=\frac95 C+32$$ you have $$\Delta F= F_2-F_1=(\frac95 C_2+32)-(\frac95 C_1+32)=\frac95 (C_2-C_1)=\frac95\Delta C$$

$$\frac C5=\frac{F-32}9\implies \frac{C_1-C_2}5=\frac{F_1-F_2}9$$

• Why are people downvoting this? Nov 15, 2014 at 8:20