1. **Problem Statement:** A homicide victim's body temperature follows Newton's Law of Cooling: $$\frac{dT}{dt} = -k(T - T_s)$$ where $T_s = 20^\circ C$ is the ambient temperature. Given: - At $t=0$ (10:30 p.m.), $T=29.5^\circ C$ - After 1.5 hours, $T=28.2^\circ C$ - Initial body temperature at time of death (unknown $t = t_0$) was $37^\circ C$ Find: (a) General equation for $T(t)$. (b) Estimate time of death. (c) Predict body temperature at 2:00 a.m. 2. **Step 1: Write the general solution of Newton's Law of Cooling** The differential equation is: $$\frac{dT}{dt} = -k(T - T_s)$$ This is a first-order linear ODE with solution: $$T(t) = T_s + (T_0 - T_s)e^{-kt}$$ where $T_0$ is the initial temperature at $t=0$. 3. **Step 2: Apply initial conditions to find $k$** At $t=0$, $T(0) = 29.5 = 20 + (T_0 - 20)e^{0} = 20 + (T_0 - 20)$ So, $$T_0 = 29.5$$ At $t=1.5$ hours, $T(1.5) = 28.2$ Using the formula: $$28.2 = 20 + (29.5 - 20)e^{-1.5k}$$ Simplify: $$28.2 - 20 = 9.5 e^{-1.5k}$$ $$8.2 = 9.5 e^{-1.5k}$$ Divide both sides: $$\frac{8.2}{9.5} = e^{-1.5k}$$ Take natural log: $$\ln\left(\frac{8.2}{9.5}\right) = -1.5k$$ Calculate: $$\ln(0.86316) = -1.5k$$ $$-0.147 = -1.5k$$ Solve for $k$: $$k = \frac{0.147}{1.5} = 0.098$$ 4. **Step 3: Write the general temperature equation** $$T(t) = 20 + 9.5 e^{-0.098 t}$$ where $t$ is hours after 10:30 p.m. 5. **Step 4: Estimate time of death** Let $t = -t_d$ be the time before 10:30 p.m. when $T = 37$ Using the formula: $$37 = 20 + 9.5 e^{-0.098 (-t_d)} = 20 + 9.5 e^{0.098 t_d}$$ Subtract 20: $$17 = 9.5 e^{0.098 t_d}$$ Divide: $$\frac{17}{9.5} = e^{0.098 t_d}$$ Take natural log: $$\ln\left(\frac{17}{9.5}\right) = 0.098 t_d$$ Calculate: $$\ln(1.789) = 0.098 t_d$$ $$0.582 = 0.098 t_d$$ Solve for $t_d$: $$t_d = \frac{0.582}{0.098} = 5.94$$ So, time of death was approximately 5.94 hours before 10:30 p.m., which is about 4:36 p.m. 6. **Step 5: Predict body temperature at 2:00 a.m.** From 10:30 p.m. to 2:00 a.m. is 3.5 hours. Calculate: $$T(3.5) = 20 + 9.5 e^{-0.098 \times 3.5}$$ Calculate exponent: $$-0.098 \times 3.5 = -0.343$$ Calculate exponential: $$e^{-0.343} = 0.709$$ Calculate temperature: $$T(3.5) = 20 + 9.5 \times 0.709 = 20 + 6.74 = 26.74^\circ C$$ **Final answers:** (a) $$T(t) = 20 + 9.5 e^{-0.098 t}$$ (b) Time of death was approximately 5.94 hours before 10:30 p.m., i.e., about 4:36 p.m. (c) Body temperature at 2:00 a.m. is approximately $$26.74^\circ C$$.