Time (sec Temperature (\"C) 30 60 90 120 150 180 210 240 270 300 330 360 390 420
ID: 578932 • Letter: T
Question
Time (sec Temperature ("C) 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510 540 570 600 83.0 81.0 78.6 77.0 75.2 73.4 72.0 70.4 69.0 67.8 66.4 65.8 65.6 65.4 65.2 65.0 64.8 64.6 64.4 64.2 a. Make a plot of tempcrature vs. time in Excel. Draw a copy below, making sure to label your axes. b. At what temperature does the slopc of the data appear to change? Make a new plot of the data, but this time use only the data points that precede the change in slope. Then, right-click on the plot and Select Data. Add another data series using the data points that occur after the change in slope. c. d. Write the formula Add a linear trendline for the first data series. Make sure to Display Equation on Chart. of the lincar trendline below e. Add a linear trendline for the second data series. Make sure to Display Equation on Chart. Write the formula of the linear trendline below f. Set the two equations equal to cach other to find the time (x-value) of intersection. Then, plug this time value back into either equation and solve for the corresponding temperature. This is the freezing point of the solution.Explanation / Answer
(a) – (e) All these parts can be answered from the plot of temperature vs time as shown below.
The slope of the curve appears to change around 65-70°C and 270-300 sec.
f) The freezing point of the solution is given by the intersection of the two linear segments. At the freezing point, the temperature, y has a fixed value and hence, the two equations are equal. Therefore,
-0.0581x + 84.228 = -0.0067x + 68.2
====> -0.0581x + 0.0067x = 68.2 – 84.228
====> -0.0514x = -16.028
====> x = (-16.028)/(-0.0514) = 311.829
The freezing point occurs at 311.829 secs. Plug x = 311.829 in either of the equations (we are using series 2 here; series 3 will give the same result) and obtain
y = -0.0581*(311.829) + 84.228 = -18.1173 + 84.228 = 66.1107 66.1
The freezing point occurs at 66.1°C (ans).
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