Answer:
A) 

B) 
C) 

Explanation:
For this case we want to create a function like this:

Where Z represent the degrees for the Z scale C the Celsius grades and tha valus a and b parameters for the model.
The boiling point of nitrogen is -195,8 °C
The melting point of iron is 1538 °C
We know the following equivalences:
-195.8 °C = 0 °Z
1538 °C = 1000 °Z
Let's say that one point its (1538C, 1000 Z) and other one is (-195.8 C, 0Z)
So then we can calculate the slope for the linear model like this:

And now for the slope we can use one point let's use for example (-195.8C, 0Z), and we have this:

And if we solve for b we got:

So then our lineal model would be:

Part A
The boiling point of water is 100C so we just need to replace in the model and see what we got:

Part B
For this case we have Z =100 and we want to solve for C, so we can do this:


Part C
For this case we know that 
And we can use the result from part B to solve for K like this:
