Question C.1.1: Water in a glass measuring cup was allowed to cool after bei...
Water in a glass measuring cup was allowed to cool after being heated to 204°F. The ambient air temperature was 70°F. The measured water temperature at various times is given in the following table.
\begin{array}{l|llllll}\text{Time}( sec ) & 0 & 120 & 240 & 360 & 480 & 600 \\\hline \text{Temperature}\left({}^{\circ}F \right) & 204 & 191 & 178 & 169 & 160 & 153 \\\text{Time (sec)}& 720 & 840 & 960 & 1080 & 1200 \\\hline \text{Temperature}\left({}^{\circ}F \right) & 147 & 141 & 137 & 132 & 127\end{array}
Obtain a functional description of the water temperature versus time.
Common sense tells us that the water temperature will eventually reach the air temperature of 70°. Thus, we first subtract 70° from the temperature data T and seek to obtain a functional description of the relative temperature, ΔT = T − 70. A plot of the relative temperature data is shown in Figure C.1.4. We note that the plot has a distinct curvature and that it does not pass through the origin. Thus, we can rule out the linear function and the power function as candidates.
To see if the data can be described by an exponential function, we plot the data on a semilog plot, which is shown in Figure C.1.5. The straight line shown can be drawn by aligning a straightedge so that it passes near most of the data points (note that this line is subjective; another person might draw a different line). The data lie close to a straight line, so we can use the exponential function to describe the relative temperature.
Using the second form of the exponential function, we can write \Delta T=b e^{m t}. Next we select two points on the straight line to find the values of b and m. The two points indicated by an asterisk were selected to minimize interpolation error because they lie near grid lines. The accuracy of the values read from the plot obviously depends on the size of the plot. Two points read from the plot are (1090, 60) and (515, 90). Using the equations developed previously to compute b and m (with t replacing x and ΔT replacing y), we have
\begin{aligned}m &=\frac{1}{1090-515}\ln \frac{60}{90}=-0.0007 \\b &=90 e^{-0.0007(515)}=129\end{aligned}
Thus, the estimated function is
\Delta T=129 e^{-0.0007 t}\quad \text{or}\quad T=\Delta T+70=129 e^{-0.0007 t}+70
where ΔT and T are in °F and time t is in seconds. The plot of ΔT versus t is shown in Figure C.1.6. From this we can see that the function provides a reasonably good description of the data. In Section C.2 we will discuss how to quantify the quality of this description.
