To use the linear regression method (least squares method):we consider the first measurement as time 1, extend the line to the y axis, where we find the intercept is 97.1. The mean value is 61.7.
With this information, we can compute the data points by x = X - Xa, where Xa is the mean value for X, and Y = Y -Ya, where Ya is the mean value for Y, and then compute values using the formula: y = (∑xy/∑x2)x
Regression line equation: Y= -1.66X + 97.1, and yields an average decline of 8.3oC per 5 minutes.
The last measurement was taken at 24.9oC, and we want to know the time needed to reach -7oC.
The temperature change will be 31.9 degrees, and we know that the cooling rate is 8.3o per 5 minutes.
So, at 45 minutes, the temperature will be 24.9 - 8.3 = 16.6
at 50 minutes, the temperature will be 16.6 - 8.3 = 8.3
at 55 minutes, the temperature will be 8.3 - 8.3 = 0
at 59.2 minutes, the temperature will be -7oC
The last value came from the fractional portion of 8.3o per 5 minutes divided by 7o.
With this information, we can compute the data points by x = X - Xa, where Xa is the mean value for X, and Y = Y -Ya, where Ya is the mean value for Y, and then compute values using the formula: y = (∑xy/∑x2)x
Regression line equation: Y= -1.66X + 97.1, and yields an average decline of 8.3oC per 5 minutes.
The last measurement was taken at 24.9oC, and we want to know the time needed to reach -7oC.
The temperature change will be 31.9 degrees, and we know that the cooling rate is 8.3o per 5 minutes.
So, at 45 minutes, the temperature will be 24.9 - 8.3 = 16.6
at 50 minutes, the temperature will be 16.6 - 8.3 = 8.3
at 55 minutes, the temperature will be 8.3 - 8.3 = 0
at 59.2 minutes, the temperature will be -7oC
The last value came from the fractional portion of 8.3o per 5 minutes divided by 7o.