The temperature at which water freezes is 0°C or 32°F. The temperature at which water boils is
100°C or 212°F. The relationship between degrees Celsius and degrees Fahrenheit is shown
through the graph below.
a.) Given, that 1°C is 33.8°F (or t1=33.8 ) and d = 1.8, write out the general formula, tn, that
represents the relationship between degrees Fahrenheit and degrees Celsius. Be sure to define
the variables.
b.) Given two points on the graph, the freezing point of water at 0°C or 32°F and the boiling
point of water at 100°C or 212°F, calculate the slope of the line. Explain how this value
relates to the general formula found in part a.
c.) What is the y-intercept of the graph? Explain how this value relates to the general formula found in part a.
4 years ago
Answered By Rohtaz S
Lets write the terminology for Temperature in Celcius and Fahrenheit as Tnc and TnF
Given:Water Melting Point :
T1C = 0oC is same as T1F= 32oF
Water Boiling Point :
T2C = 100oC is same as T2F = 212oF
Lets plot Degree Celcius (TnC) on X- axis and (TnF) Fahrenheit on y-axis.
As we have two points, we can write down equation of a straight line as following:
$\frac{\left(TnF-T1F\right)}{\left(TnC-T1C\right)}=\frac{T2F-T1F}{T2C-T1C}$(TnF−T1F)(TnC−T1C) =T2F−T1FT2C−T1C Insert values for T at Melting and Boiling Point=> $\frac{TnF-32}{TnC-0}=\frac{212-32}{100-0}$TnF−32TnC−0 =212−32100−0 Simplify and Solve the equation
=> $TnF-32=\frac{9}{5}TnC$TnF−32=95 TnC
or $TnF=\frac{9}{5}TnC+32$TnF=95 TnC+32 (General Equation)
The general form of a straight line can be written down as follows
$y=mx+c$y=mx+c
where m is the slope and c is the y intercept, which in simple terms is the point on y axis when x is equal to 0
so based on the general form, $m=\frac{9}{5}$m=95 and
y intercept = 32
Also, the general equation can be validated for T1C = 1oC and T1F = 33.8oF
lets insert T1C in the general equation:
$T1F=\frac{9}{5}\left(1\right)+32$T1F=95 (1)+32
=> $T1F=1.8\left(1\right)+32=33.8$T1F=1.8(1)+32=33.8