WARNING! This is a long answer.

Assume that you are doing a coffee cup calorimeter experiment in which you have to measure ##ΔT##. You measure the temperatures over a period of time and plot a graph.

But the maximum temperature is not that shown on the graph, because the calorimeter is not a perfect insulator. It is constantly losing heat to the room.

The teacher asks you to use Microsoft Excel to extrapolate the data back to ##”time” = 0##.

Here’s how I do it using Excel 2010 for Windows. You may have to modify for your version.

First, watch Mr. Pauller’s video on .

Enter your data into Columns A and B of the Excel spreadsheet.

Assume that your data are:

Highlight the data from A1…B13. Click on **Insert Charts**.

Click the small triangle under **Scatter**. Select the first option (the graph with just points and no lines).

You should get a graph that looks like this.

The peak at (5.0, 30.0) has the maximum temperature. The points from 5 min to 12 min are in a sloped but linear region.

Go back to the spreadsheet and copy these values into column C directly across from their values in column B (as in the data table above).

Right-click in a blank portion of the **Plot Area.** Click on **Select Data…**

The **Select Data Source** window will open. Click the **Add** button.

The **Edit Series** window will open. Click in the **Series X Values** space and highlight cells A2…A13.

Delete the entry in the **Series Y Values** space and highlight cells C2…C13. Click **OK**, then **OK** again.

A new series of points appears on your graph.

Delete the entries in cells B7…B13, since they are now duplicated in C7… C13.

Right-click on a point in **Series 2**. Click on **Add Trendline…**

The **Format Trendline** window will open. Choose **Linear Trend/Regression**. Click **Display Equation on Chart**. Then **Close**.

If necessary, move the equation to a clear area on the chart.

The trend line appears and extends back to intersect the ##y##-axis at about 32.5 °C.

You can also set ##x = 0## in the regression equation ##y = 0.5 x + 32.5##. Again you get 32.5 °C for the extrapolated temperature.

Here is the source document for the original answer.