1. **Problem statement:** We have a cooling process of a hot liquid in an environment at constant temperature 20 °C. The rate of temperature decrease is given by the function $$a(t) = -69 \cdot k \cdot e^{-k \cdot t}$$ where $a(t)$ is the rate of temperature change at time $t$. We need to answer: (a) What type of cooling process is this? Explain. (b) What was the initial temperature at $t=0$? (c) Given that at $t=3$ minutes the temperature is 73 °C, find the time when the temperature decrease rate first becomes less than 1 degree per minute. 2. **Type of cooling process (a):** The rate of temperature change is proportional to an exponential decay function $e^{-k t}$, which suggests Newton's law of cooling. This law states that the temperature difference between the object and the environment decreases exponentially over time. The general form of Newton's law of cooling is: $$T(t) = T_{env} + (T_0 - T_{env}) e^{-k t}$$ where $T_{env} = 20$ °C is the environment temperature, $T_0$ is the initial temperature, and $k$ is a positive constant. 3. **Initial temperature (b):** The rate of temperature change $a(t)$ is the derivative of $T(t)$: $$a(t) = \frac{dT}{dt} = -k (T_0 - T_{env}) e^{-k t}$$ Given $a(t) = -69 k e^{-k t}$, we identify: $$-k (T_0 - 20) e^{-k t} = -69 k e^{-k t}$$ Dividing both sides by $-k e^{-k t}$ (which is nonzero for $t \geq 0$): $$\cancel{-k} (T_0 - 20) \cancel{e^{-k t}} / \cancel{-k} \cancel{e^{-k t}} = \cancel{-69} \cancel{k} \cancel{e^{-k t}} / \cancel{-k} \cancel{e^{-k t}}$$ which simplifies to: $$T_0 - 20 = 69$$ Therefore, $$T_0 = 89 \text{ °C}$$ 4. **Find $k$ using temperature at $t=3$ (c):** We know: $$T(3) = 20 + (89 - 20) e^{-3k} = 73$$ Simplify: $$20 + 69 e^{-3k} = 73$$ Subtract 20: $$69 e^{-3k} = 53$$ Divide both sides by 69: $$e^{-3k} = \frac{53}{69}$$ Take natural logarithm: $$-3k = \ln\left(\frac{53}{69}\right)$$ $$k = -\frac{1}{3} \ln\left(\frac{53}{69}\right)$$ Calculate numeric value: $$\ln\left(\frac{53}{69}\right) = \ln(0.7681) \approx -0.264$$ So, $$k = -\frac{1}{3} (-0.264) = 0.088$$ 5. **Find time when temperature decrease rate is less than 1 °C/min:** Recall: $$a(t) = -69 k e^{-k t}$$ We want the magnitude of $a(t)$ to be less than 1: $$|a(t)| = 69 k e^{-k t} < 1$$ Divide both sides by $69 k$: $$e^{-k t} < \frac{1}{69 k}$$ Take natural logarithm: $$-k t < \ln\left(\frac{1}{69 k}\right)$$ Multiply both sides by $-1$ (reverse inequality): $$k t > -\ln\left(\frac{1}{69 k}\right) = \ln(69 k)$$ Solve for $t$: $$t > \frac{\ln(69 k)}{k}$$ Calculate numeric value: $$69 k = 69 \times 0.088 = 6.072$$ $$\ln(6.072) \approx 1.804$$ Therefore, $$t > \frac{1.804}{0.088} \approx 20.5 \text{ minutes}$$ **Final answers:** (a) The cooling process follows Newton's law of cooling, where temperature difference decreases exponentially. (b) Initial temperature $T_0 = 89$ °C. (c) The temperature decrease rate first becomes less than 1 °C/min after approximately 20.5 minutes.