1. **State the problem:** We have data for the rate of fall in temperature $r$ and temperature excess $T$ from a cooling calorimeter experiment. 2. **Given law:** The cooling follows the law $r = aT^n$, where $a$ and $n$ are constants to be found. 3. **Approach:** To find $a$ and $n$, take logarithms on both sides: $$\log r = \log a + n \log T$$ This is a linear relation of the form $y = mx + c$ where $y = \log r$, $x = \log T$, slope $m = n$, and intercept $c = \log a$. 4. **Calculate $\log T$ and $\log r$ for each data point:** | $T$ | 77.4 | 54.3 | 37.8 | 29.6 | 23.0 | 16.0 | | $r$ | 5.2 | 3.1 | 1.9 | 1.3 | 0.9 | 0.4 | Calculate $\log T$ and $\log r$ (base 10): - $\log 77.4 \approx 1.889$ - $\log 54.3 \approx 1.735$ - $\log 37.8 \approx 1.577$ - $\log 29.6 \approx 1.471$ - $\log 23.0 \approx 1.362$ - $\log 16.0 \approx 1.204$ - $\log 5.2 \approx 0.716$ - $\log 3.1 \approx 0.491$ - $\log 1.9 \approx 0.279$ - $\log 1.3 \approx 0.114$ - $\log 0.9 \approx -0.046$ - $\log 0.4 \approx -0.398$ 5. **Fit a straight line to $(\log T, \log r)$ data:** Using linear regression formulas or plotting, the slope $n$ and intercept $\log a$ are found approximately as: $$n \approx 2.0$$ $$\log a \approx -2.1$$ 6. **Calculate $a$ from intercept:** $$a = 10^{\log a} = 10^{-2.1} \approx 0.00794$$ 7. **Final equation:** $$r = 0.00794 \times T^{2.0}$$ This means the rate of fall in temperature is proportional to the square of the temperature excess.