Let’s break down the key steps and solutions for each part of the Newton’s Law of Cooling project as described in the image. --- ### **B. Derivation of Exponential Decay Equation** Newton's Law of Cooling: \[ \frac{dT}{dt} = k [T(t) - T_a] \] where: - \( T(t) \): Temperature at time \( t \) - \( T_a \): Ambient temperature - \( k \): Growth (decay) constant --- #### **(i) Separable Variables Method** 1. **Rewrite the differential equation:** \[ \frac{dT}{dt} = k (T - T_a) \] 2. **Separate variables:** \[ \frac{1}{T-T_a} dT = k dt \] 3. **Integrate both sides:** \[ \int \frac{1}{T-T_a} dT = \int k dt \] \[ \ln|T-T_a| = kt + C \] 4. **Exponentiate both sides:** \[ |T-T_a| = e^{kt+C} = Ae^{kt} \] Where \( A = e^C \). 5. **Drop the absolute value (since temperature difference can be positive or negative):** \[ T - T_a = Ae^{kt} \] \[ T(t) = T_a + Ae^{kt} \] 6. **To find \( A \) using initial condition \( T(0) = T_i \):** \[ T_i = T_a + A \implies A = T_i - T_a \] \[ \boxed{T(t) = T_a + (T_i - T_a) e^{kt}} \] --- #### **(ii) Integration Factor Method** 1. **Write in standard linear ODE form:** \[ \frac{dT}{dt} - kT = -kT_a \] 2. **Find the integrating factor (\(\mu(t)\)):** \[ \mu(t) = e^{-\int k dt} = e^{-kt} \] 3. **Multiply through by integrating factor:** \[ e^{-kt} \frac{dT}{dt} - k e^{-kt} T = -kT_a e^{-kt} \] Left side is derivative: \[ \frac{d}{dt}[T e^{-kt}] = -kT_a e^{-kt} \] 4. **Integrate both sides:** \[ \int \frac{d}{dt}[T e^{-kt}] dt = \int -kT_a e^{-kt} dt \] \[ T e^{-kt} = T_a e^{-kt} + C \] 5. **Solve for \( T \):** \[ T = T_a + C e^{kt} \] Using initial condition \( T(0) = T_i \): \[ T_i = T_a + C \implies C = T_i - T_a \] \[ \boxed{T(t) = T_a + (T_i - T_a) e^{kt}} \] --- ### **C. Using the General Solution** #### **(i) Particular Solution** Plug in the given data to find \( k \) or any unknowns using \( T(t) = T_a + (T_i - T_a) e^{kt} \). #### **(ii) Initial Temperature \( T_i \)** Given \( T(t) \) at a specific time (not at \( t=0 \)), use: \[ T(t) = T_a + (T_i - T_a) e^{kt} \] Solve for \( T_i \). --- ### **D. Time to Cool to Half Initial Temperature** Set \( T(t_{1/2}) = T_a + \frac{1}{2}(T_i - T_a) \): \[ T_a + \frac{1}{2}(T_i - T_a) = T_a + (T_i - T_a) e^{kt_{1/2}} \] \[ \frac{1}{2} = e^{kt_{1/2}} \] \[ kt_{1/2} = \ln\left(\frac{1}{2}\right) = -\ln 2 \] \[ \boxed{t_{1/2} = -\frac{\ln 2}{k}} \] --- ### **E. Mathematical Proof of Half-Time** Let \( T_d(t) = T(t) - T_a \): \[ T_d(t) = (T_i - T_a) e^{kt} \] For half-time, \( T_d(t_{1/2}) = \frac{1}{2}(T_i - T_a) \): \[ \frac{1}{2}(T_i - T_a) = (T_i - T_a) e^{kt_{1/2}} \] \[ \frac{1}{2} = e^{kt_{1/2}} \] \[ kt_{1/2} = -\ln 2 \] \[ \boxed{t_{1/2} = -\frac{\ln 2}{k}} \] --- ### **F. Doubling Time for Exponential Growth** For growth: \( T(t) = T_a + (T_i - T_a) e^{kt} \), but for exponential growth (e.g., population): Doubling time (\( t_{double} \)): \[ 2A = A e^{kt_{double}} \] \[ 2 = e^{kt_{double}} \] \[ kt_{double} = \ln 2 \] \[ \boxed{t_{double} = \frac{\ln 2}{k}} \] --- ### **G. Similarity & Difference: Doubling Time vs. Half-Life** **Similarity:** Both are derived from the same exponential model and use the factor \( \ln 2 \). **Difference:** - *Half-life* is the time to decrease to half the initial value (decay, negative \( k \)). - *Doubling time* is the time to increase to twice the initial value (growth, positive \( k \)). --- #### **Summary Table** | Part | Final Formula/Answer | |------|---------------------| | B(i), B(ii) | \( T(t) = T_a + (T_i - T_a) e^{kt} \) | | D, E | \( t_{1/2} = -\frac{\ln 2}{k} \) | | F | \( t_{double} = \frac{\ln 2}{k} \) | | G | Both use \( \ln 2 \), but one halves (decay), one doubles (growth) | --- If you need a worked numerical example or further steps, let me know!

Certainly! Below is a detailed step-by-step solution with calculations for each part of the project, including the derivations, formulas, and final answers. --- ## **Part B: Derivation of Exponential Decay Equation** Given Newton's Law of Cooling: \[ \frac{dT}{dt} = k (T - T_a) \] where: - \( T(t) \): Temperature at time \( t \) - \( T_a \): Ambient temperature (constant) - \( k \): Cooling constant (negative for cooling) ### **(i) Separable Variables Method** **Step 1:** Rearrange the differential equation: \[ \frac{dT}{dt} = k (T - T_a) \] \[ \Rightarrow \frac{1}{T - T_a} dT = k dt \] **Step 2:** Integrate both sides: \[ \int \frac{1}{T - T_a} dT = \int k dt \] \[ \ln|T - T_a| = kt + C \] **Step 3:** Exponentiate: \[ |T - T_a| = e^{kt + C} = e^{C} e^{kt} = A e^{kt} \] where \( A = e^{C} \). **Step 4:** Write the general solution: \[ T(t) = T_a + A e^{kt} \] **Step 5:** Use initial condition \( T(0) = T_i \): \[ T_i = T_a + A e^{k \times 0} = T_a + A \] \[ \Rightarrow A = T_i - T_a \] **Final exponential decay equation:** \[ \boxed{ \textbf{(1)} \quad T(t) = T_a + (T_i - T_a) e^{kt} } \] --- ### **(ii) Integration Factor Method** **Step 1:** Rewrite the differential equation in linear form: \[ \frac{dT}{dt} - k T = -k T_a \] **Step 2:** Find the integrating factor: \[ \mu(t) = e^{-\int k dt} = e^{-kt} \] **Step 3:** Multiply entire equation by \(\mu(t)\): \[ e^{-kt} \frac{dT}{dt} - k e^{-kt} T = -k T_a e^{-kt} \] which simplifies to: \[ \frac{d}{dt} [T e^{-kt}] = -k T_a e^{-kt} \] **Step 4:** Integrate both sides: \[ T e^{-kt} = \int -k T_a e^{-kt} dt + C \] The integral: \[ \int -k T_a e^{-kt} dt \] Let’s compute it: \[ \int -k T_a e^{-kt} dt \] **Substitution:** \[ u = -kt \Rightarrow du = -k dt \Rightarrow dt = -\frac{1}{k} du \] Rewriting: \[ \int -k T_a e^{u} \left(-\frac{1}{k} \right) du = T_a \int e^{u} du = T_a e^{u} + D \] Back to \( t \): \[ T_a e^{-kt} + D \] Thus, \[ T e^{-kt} = T_a e^{-kt} + C \] **Step 5:** Solve for \( T(t) \): \[ T(t) = T_a + C e^{kt} \] Apply the initial condition \( T(0) = T_i \): \[ T_i = T_a + C \Rightarrow C = T_i - T_a \] **Final solution:** \[ \boxed{ \textbf{(2)} \quad T(t) = T_a + (T_i - T_a) e^{kt} } \] --- ## **Part C: Determine the Particular Solution and Initial Temperature** Given data (assumed for example): | Time \( t \) | Temperature \( T(t) \) | |--------------|-------------------------| | 0 min | \( T_i \) (initial) | | 5 min | \( T_{5} \) | Suppose from measurements: - \( T_a = 25^\circ C \) - \( T(0) = 80^\circ C \) - \( T(5) = 50^\circ C \) ### **(i) Find \( T_i \):** Since \( T(0) = 80^\circ C \), initial temperature: \[ \boxed{ T_i = 80^\circ C } \] ### **(ii) Find \( k \) using data at \( t=5 \) min:** Using the exponential model: \[ T(5) = T_a + (T_i - T_a) e^{k \times 5} \] \[ 50 = 25 + (80 - 25) e^{5k} \] \[ 50 - 25 = 55 e^{5k} \] \[ 25 = 55 e^{5k} \] \[ e^{5k} = \frac{25}{55} = \frac{5}{11} \approx 0.4545 \] \[ 5k = \ln(0.4545) \approx -0.790 \] \[ k = \frac{-0.790}{5} \approx -0.158 \text{ min}^{-1} \] --- ## **Part D: Time to Cool to Half of Initial Temperature Difference** Initial temperature difference: \[ \Delta T_0 = T_i - T_a = 80 - 25 = 55^\circ C \] Target temperature: \[ T_{1/2} = T_a + \frac{1}{2} \Delta T_0 = 25 + 27.5 = 52.5^\circ C \] Set in the exponential equation: \[ 52.5 = 25 + 55 e^{k t_{1/2}} \] \[ 52.5 - 25 = 55 e^{k t_{1/2}} \] \[ 27.5 = 55 e^{k t_{1/2}} \] \[ e^{k t_{1/2}} = \frac{27.5}{55} = 0.5 \] \[ k t_{1/2} = \ln 0.5 = -0.6931 \] Using \( k = -0.158 \): \[ t_{1/2} = \frac{-0.6931}{-0.158} \approx 4.39 \text{ minutes} \] **Final answer:** \[ \boxed{ t_{1/2} \approx \mathbf{4.39 \text{ minutes}} } \] --- ## **Part E: Mathematical Proof that \( t_{1/2} = -\frac{\ln 2}{k} \)** Starting with the general decay model: \[ T_d(t) = T(t) - T_a = (T_i - T_a) e^{k t} \] At half-life: \[ T_d(t_{1/2}) = \frac{1}{2} T_d(0) = \frac{1}{2}(T_i - T_a) \] \[ \Rightarrow \frac{1}{2} (T_i - T_a) = (T_i - T_a) e^{k t_{1/2}} \] \[ \Rightarrow \frac{1}{2} = e^{k t_{1/2}} \] \[ \Rightarrow k t_{1/2} = \ln \left( \frac{1}{2} \right) = -\ln 2 \] \[ \boxed{ t_{1/2} = - \frac{\ln 2}{k} } \] --- ## **Part F: Exponential Doubling Time** For growth (assuming \( T(t) = T_a + (T_i - T_a) e^{k t} \) with positive \( k \)), the doubling time \( t_{double} \) satisfies: \[ T(t_{double}) = 2 T_i \quad \text{(if initial temperature is \( T_i \))} \] or, in terms of the temperature difference: \[ T_d(t) = (T_i - T_a) e^{k t} \] Set: \[ (T_i - T_a) e^{k t_{double}} = 2 (T_i - T_a) \] \[ e^{k t_{double}} = 2 \] \[ k t_{double} = \ln 2 \] \[ \boxed{ t_{double} = \frac{\ln 2}{k} } \] --- ## **Part G: Similarity and Difference between Half-Time and Doubling Time** - **Similarity:** Both are derived from the exponential function involving \( e^{k t} \) and involve the constant \( \ln 2 \). - **Difference:** - *Half-life* refers to the time for the quantity to decrease to half its initial value (decay, negative \( k \)), and is \( t_{1/2} = - \frac{\ln 2}{k} \). - *Doubling time* refers to the time for the quantity to double (growth, positive \( k \)), and is \( t_{double} = \frac{\ln 2}{k} \). --- ## **Summary of Final Results:** | Part | Final Expression / Numerical Result | |---------|-------------------------------------| | B(i), B(ii) | \( T(t) = T_a + (T_i - T_a) e^{kt} \) | | C | \( T_i = 80^\circ C \), \( k \approx -0.158 \text{ min}^{-1} \) | | D | \( t_{1/2} \approx 4.39 \text{ minutes} \) | | E | Proven: \( t_{1/2} = - \frac{\ln 2}{k} \) | | F | \( t_{double} = \frac{\ln 2}{k} \) | | G | Same form involving \( \ln 2 \); half-life decreases with negative \( k \), doubling time increases with positive \( k \). | --- **Note:** Replace the assumed data with your actual measurements to perform precise calculations for your project. --- If you have specific data points or need further clarifications, feel free to ask!

Certainly! Here are all the final answers summarized: ### **Part B: Exponential Decay Equation** \[ \boxed{ T(t) = T_a + (T_i - T_a) e^{k t} } \] ### **Part C: Initial Temperature and Particular Solution** - Initial temperature: \[ T_i = \text{Given or measured value} \] - Decay constant (example calculation): \[ k \approx -0.158 \text{ min}^{-1} \] ### **Part D: Time to Cool to Half Initial Temperature** \[ \boxed{ t_{1/2} = - \frac{\ln 2}{k} \approx 4.39 \text{ minutes} } \] ### **Part E: Mathematical Proof of Half-Time** \[ t_{1/2} = - \frac{\ln 2}{k} \] ### **Part F: Exponential Doubling Time** \[ \boxed{ t_{double} = \frac{\ln 2}{k} } \] ### **Part G: Similarity and Difference** - **Similarity:** Both involve \( \ln 2 \) and exponential functions. - **Difference:** Half-life pertains to decay (reducing to half), doubling time pertains to growth (increasing to double). --- Use your specific data to compute \(k\) and substitute accordingly for precise results.