Question

Solve the differential equation for Newton's Law of Cooling to find the temperature in the following cases. Then answer any additional questions. A pot of boiling soup \(\left(100^{\circ} \mathrm{C}\right)\) is put in a cellar with a temperature of \(10^{\circ} \mathrm{C}\). After 30 minutes, the soup has cooled to \(80^{\circ} \mathrm{C}\). When will the temperature of the soup reach \(30^{\circ} \mathrm{C} ?\)

Short answer

Answer: The temperature of the soup will reach 30°C at the time \(t = -\frac{30\ln\left(\frac{2}{9}\right)}{\ln\left(\frac{7}{9}\right)}\) minutes.

Step-by-step solution

Write down Newton's Law of Cooling equation

Newton's Law of Cooling can be represented by the following differential equation: \(\frac{dT}{dt} = -k\left(T - T_{ambient}\right)\) Where \(T(t)\) is the temperature of the object at time \(t\), \(T_{ambient}\) is the ambient temperature, and \(k\) is a positive proportional constant.

Solve the differential equation

To solve the first-order linear differential equation, separate the variables and integrate both sides: \(\int\frac{dT}{T - T_{ambient}} = -k\int dt\) After integrating, we get: \(\ln\left|T - T_{ambient}\right| = -kt + C\) Where \(C\) is the constant of integration. Now, remove the natural logarithm with an exponential function: \(T - T_{ambient} = Ae^{-kt}\) Where \(A = e^C\) is another constant.

Apply initial condition and find A

We know that at the beginning, \(t = 0\) and \(T(0) = 100^\circ C\). Substitute these values into the equation and solve for \(A\): \(100 - 10 = Ae^{-k\cdot 0}\) \(A = 90\) Now, our equation becomes: \(T - 10 = 90e^{-kt}\)

Apply the condition after 30 minutes and find k

After 30 minutes, the temperature of the soup is \(80^\circ C\). Substitute \(t = 30\) and \(T(30) = 80\) into the equation and solve for \(k\): \(80 - 10 = 90e^{-k\cdot 30}\) Divide by 90 and simplify: \(\frac{70}{90} = e^{-30k}\) Take the natural logarithm of both sides: \(\ln\left(\frac{7}{9}\right) = -30k\) Solve for \(k\): \(k = -\frac{1}{30}\ln\left(\frac{7}{9}\right)\)

Set the equation for T=30 and solve for t

We want to find when the temperature reaches \(30^\circ C\). Set \(T = 30\) in our equation and solve for \(t\): \(30 - 10 = 90e^{-\left(-\frac{1}{30}\ln\left(\frac{7}{9}\right)\right)t}\) Simplify and solve for \(t\): \(20 = 90e^{-\left(\frac{1}{30}\ln\left(\frac{7}{9}\right)\right)t}\) Divide by 90: \(\frac{2}{9} = e^{-\left(\frac{1}{30}\ln\left(\frac{7}{9}\right)\right)t}\) Take the natural logarithm of both sides: \(\ln\left(\frac{2}{9}\right) = -\left(\frac{1}{30}\ln\left(\frac{7}{9}\right)\right)t\) Solve for \(t\): \(t= -\frac{30\ln\left(\frac{2}{9}\right)}{\ln\left(\frac{7}{9}\right)}\) Now, we have found the time when the temperature of the soup reaches \(30^\circ C\).

Key concepts

Newton's Law of Cooling

Newton's Law of Cooling is a principle used to describe the cooling process of an object in an environment with a different temperature. This law states that the rate of heat loss

• is proportional to the temperature difference between the object and its surroundings.
• It is formalized with the differential equation: \( \frac{dT}{dt} = -k(T - T_{ambient}) \).

Here:

• \(T(t)\) is the temperature of the object at time \(t\).
• \(T_{ambient}\) is the constant ambient temperature.
• \(k\) is a positive constant of proportionality, which reflects how quickly the object cools in its environment.

To apply Newton's Law of Cooling effectively, the key is understanding the involved variables and how they relate to the environment. The formula helps predict how long it will take an object to cool to a certain temperature. By solving this differential equation, students can understand real-world applications, such as predicting cooling times and managing thermal processes.

Initial Value Problem

An Initial Value Problem (IVP) refers to a type of problem in differential equations where the solution must satisfy a given initial condition. The initial condition is usually provided along with the differential equation.
In our context, the initial value is the starting temperature of the soup (\(100^{\circ} C\)).
An IVP typically includes:

• A differential equation, outlining how variables like temperature change with time.
• An initial condition, which is the value of the function and its derivatives at a particular point, often when \(t = 0\).

IVPs are fundamental in finding a specific solution to a differential equation that fits a real-world scenario. The initial value helps to find constants in the solution, making it specific to the context at hand.

Exponential Functions

Exponential functions describe mathematical models where a quantity grows or decays at a rate proportional to its current value. They play a crucial role in representing solutions to equations like Newton's Law of Cooling.
In such scenarios, the relationship appears as:

• \(T(t) - T_{ambient} = Ae^{-kt}\)

Here, \(A\) and \(-kt\) represent constants derived from initial conditions or system-specific parameters.
Exponential functions capture both growth and decay:

• The negative sign in the exponent in \(e^{-kt}\) indicates a decay process: the temperature of the soup decreases over time.
• The function approaches zero as \(t\) increases, meaning the difference between the object's temperature and the environment decreases.

Understanding exponential functions is essential to model and solve problems involving temperature changes, biological decay, and financial growth. They illustrate how values decrease exponentially for scenarios involving cooling, such as with the soup cooling over time.