Question

So the problem is to create a plot of points and plot them using only 'o' and no lines then using the polyfit and polyval functions, find a first order equation to curve fit the points (and plot it on the existing plot).

I got that part and have the following code

clear
clc
%define measured data
temp = [250 300 340 400 460 500 540 600 660 700 750 810 930 1000 1120]
HC= [.791 .846 .895 .939 .978 1.014 1.046 1.075 1.102 1.126 1.148 1.169 1.204 1.234 1.328]
%calculate new HC values using polyfit and polyval
p=polyfit(temp,HC,1)
y1=polyval(p,temp)
%plot original data & 1st order curve
plot(temp,HC,'o')
hold on
plot(temp, y1)

The next part though is giving me trouble: Once you have found the slope and intercept, use those values to find the Heat capacity when the temperature is 800 and display the answer.

Answer 1

Executable code:

   clear

clc

%define measured data

temp = [250 300 340 400 460 500 540 600 660 700 750 810 930

1000 1120];

HC= [.791 .846 .895 .939 .978 1.014 1.046 1.075 1.102 1.126

1.148 1.169

1.204 1.234 1.328];

%calculate new HC values using polyfit and polyval

p=polyfit(temp,HC,1)

y1=polyval(p,temp);

%plot original data & 1st order curve

plot(temp,HC,'o')

hold on

plot(temp, y1)

%heat capacity when temperature is 800, where slope and

%intercept are 0.0006 and 0.7075 .it is in variable p

HC_800=(0.0006*800)+0.7075