Answer:
A) ![Z = 0.577 C +112.931](https://tex.z-dn.net/?f=%20Z%20%3D%200.577%20C%20%2B112.931)
![Z = 0.577*(100) +112.931=170.631 Z](https://tex.z-dn.net/?f=%20Z%20%3D%200.577%2A%28100%29%20%2B112.931%3D170.631%20Z)
B) ![C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C](https://tex.z-dn.net/?f=%20C%3D%5Cfrac%7BZ-112.931%7D%7B0.577%7D%3D%20%5Cfrac%7B100-112.931%7D%7B0.577%7D%3D-22.41%20C)
C) ![K = C +273.15](https://tex.z-dn.net/?f=%20K%20%3D%20C%20%2B273.15)
![K = -22.41 +273.15 =250.739 K](https://tex.z-dn.net/?f=%20K%20%3D%20-22.41%20%2B273.15%20%3D250.739%20K)
Explanation:
For this case we want to create a function like this:
![Z = a C + b](https://tex.z-dn.net/?f=%20Z%20%3D%20a%20C%20%2B%20b)
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:
![a = \frac{z_2 -z_1}{c_2 -c_1}= \frac{1000 Z- 0Z}{1538C -(-195.8 C)}=\frac{1000 Z}{1733.8 C}=0.577 \frac{Z}{C}](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7Bz_2%20-z_1%7D%7Bc_2%20-c_1%7D%3D%20%5Cfrac%7B1000%20Z-%200Z%7D%7B1538C%20-%28-195.8%20C%29%7D%3D%5Cfrac%7B1000%20Z%7D%7B1733.8%20C%7D%3D0.577%20%5Cfrac%7BZ%7D%7BC%7D)
And now for the slope we can use one point let's use for example (-195.8C, 0Z), and we have this:
![0 = 0.577 (-195.8) + b](https://tex.z-dn.net/?f=%200%20%3D%200.577%20%28-195.8%29%20%2B%20b)
And if we solve for b we got:
![b = 0.577*195.8 =112.931 Z](https://tex.z-dn.net/?f=%20b%20%3D%200.577%2A195.8%20%3D112.931%20Z)
So then our lineal model would be:
![Z = 0.577 C +112.931](https://tex.z-dn.net/?f=%20Z%20%3D%200.577%20C%20%2B112.931)
Part A
The boiling point of water is 100C so we just need to replace in the model and see what we got:
![Z = 0.577*(100) +112.931=170.631 Z](https://tex.z-dn.net/?f=%20Z%20%3D%200.577%2A%28100%29%20%2B112.931%3D170.631%20Z)
Part B
For this case we have Z =100 and we want to solve for C, so we can do this:
![Z-112.931 = 0.577 C](https://tex.z-dn.net/?f=%20Z-112.931%20%3D%200.577%20C)
![C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C](https://tex.z-dn.net/?f=%20C%3D%5Cfrac%7BZ-112.931%7D%7B0.577%7D%3D%20%5Cfrac%7B100-112.931%7D%7B0.577%7D%3D-22.41%20C)
Part C
For this case we know that ![K = C +273.15](https://tex.z-dn.net/?f=%20K%20%3D%20C%20%2B273.15)
And we can use the result from part B to solve for K like this:
![K = -22.41 +273.15 =250.739 K](https://tex.z-dn.net/?f=%20K%20%3D%20-22.41%20%2B273.15%20%3D250.739%20K)