How to Solve Newton's Law of Cooling Problems Using Differential Equations

📚 Understanding Newton's Law of Cooling

Newton's Law of Cooling describes the rate at which an object's temperature changes relative to its surroundings. Essentially, a hot object cools faster in a cooler environment, and a cold object warms up faster in a warmer environment. This principle is widely used in various fields, from forensic science to engineering.

📜 Historical Context

Sir Isaac Newton formulated this law in the late 17th century. While not as famous as his laws of motion or universal gravitation, Newton's Law of Cooling has significant practical applications. It was initially based on empirical observations of how heated objects cool down, and it laid groundwork for further studies in thermodynamics.

🌡️ Key Principles and Formula

The law states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e., the temperature of its surroundings). Mathematically, it can be expressed as:

$\frac{dT}{dt} = -k(T - T_s)$

Where:

• 🌡️ $T$ is the temperature of the object at time $t$.
• 🕰️ $t$ is the time.
• 🔄 $T_s$ is the surrounding temperature (ambient temperature).
• ❄️ $k$ is a positive constant that depends on the properties of the object (e.g., its material, shape, and surface area).

📝 Solving Newton's Law of Cooling Problems

To solve problems using Newton's Law of Cooling, follow these steps:

• 🔍 Identify the given values: $T_s$, initial temperature $T(0)$, and any other temperature $T$ at a specific time $t$.
• ✍️ Set up the differential equation: $\frac{dT}{dt} = -k(T - T_s)$.
• ➗ Solve the differential equation. This is a separable equation, and its general solution is: $T(t) = T_s + (T(0) - T_s)e^{-kt}$.
• 🧩 Use the given information to find the constant $k$.
• 📈 Once you have $k$, you can use the equation to find the temperature at any time $t$, or the time it takes to reach a specific temperature.

☕ Real-world Examples

Example 1: Cooling Coffee

A cup of coffee is initially at 90°C and is placed in a room at 20°C. After 20 minutes, the coffee has cooled to 60°C. What is the temperature of the coffee after 40 minutes?

1. ✍️ $T_s = 20°C$, $T(0) = 90°C$, $T(20) = 60°C$
2. ➗ $T(t) = 20 + (90 - 20)e^{-kt} = 20 + 70e^{-kt}$
3. 🧩 $60 = 20 + 70e^{-20k} \Rightarrow e^{-20k} = \frac{4}{7} \Rightarrow k = -\frac{1}{20}\ln(\frac{4}{7}) \approx 0.02798$
4. 📈 $T(40) = 20 + 70e^{-40(0.02798)} \approx 42.86°C$

Example 2: Forensic Science

A body is found in a room with a constant temperature of 22°C. The body's temperature is initially measured at 30°C, and after one hour, it has cooled to 28°C. Assuming normal body temperature at the time of death was 37°C, how long ago did the person die?

1. ✍️ $T_s = 22°C$, $T(0) = 30°C$, $T(1) = 28°C$
2. ➗ $T(t) = 22 + (30 - 22)e^{-kt} = 22 + 8e^{-kt}$
3. 🧩 $28 = 22 + 8e^{-k} \Rightarrow e^{-k} = \frac{6}{8} \Rightarrow k = - \ln(\frac{3}{4}) \approx 0.2877$
4. 📈 To find the time of death ($t_d$), we solve $37 = 22 + 8e^{-kt_d} \Rightarrow e^{-kt_d} = \frac{15}{8} \Rightarrow t_d = -\frac{1}{k}\ln(\frac{15}{8}) \approx -\frac{\ln(1.875)}{-0.2877} \approx 2.32 \text{ hours}$

💡 Tips for Success

• 🧪 Always double-check your units to ensure consistency (e.g., time in minutes or hours, temperature in Celsius or Fahrenheit).
• 📐 Pay close attention to the initial conditions and ambient temperature, as they are crucial for solving the differential equation.
• 🔢 Practice with various problems to become comfortable with the different scenarios and applications of Newton's Law of Cooling.

🔑 Conclusion

Newton's Law of Cooling provides a practical way to understand and predict temperature changes in various scenarios. By understanding the underlying principles and practicing problem-solving, you can master this concept and apply it effectively in real-world situations.