Respuesta :
Answer:
A) [tex] Z = 0.577 C +112.931[/tex]
[tex] Z = 0.577*(100) +112.931=170.631 Z[/tex]
B) [tex] C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C[/tex]
C) [tex] K = C +273.15[/tex]
[tex] K = -22.41 +273.15 =250.739 K[/tex]
Explanation:
For this case we want to create a function like this:
[tex] Z = a C + b[/tex]
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:
[tex] 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}[/tex]
And now for the slope we can use one point let's use for example (-195.8C, 0Z), and we have this:
[tex] 0 = 0.577 (-195.8) + b[/tex]
And if we solve for b we got:
[tex] b = 0.577*195.8 =112.931 Z[/tex]
So then our lineal model would be:
[tex] Z = 0.577 C +112.931[/tex]
Part A
The boiling point of water is 100C so we just need to replace in the model and see what we got:
[tex] Z = 0.577*(100) +112.931=170.631 Z[/tex]
Part B
For this case we have Z =100 and we want to solve for C, so we can do this:
[tex] Z-112.931 = 0.577 C[/tex]
[tex] C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C[/tex]
Part C
For this case we know that [tex] K = C +273.15[/tex]
And we can use the result from part B to solve for K like this:
[tex] K = -22.41 +273.15 =250.739 K[/tex]
