Question

Solve the differential equation for Newton's Law of Cooling to find the temperature function 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 time it takes for the soup to reach \(30^{\circ} \mathrm{C}\) is approximately \(\frac{-30\ln\left(\frac{2}{9}\right)}{\ln(\frac{9}{7})}\) minutes.

Step-by-step solution

Separation of variables

Separate variables to integrate both sides of the equation: \(\frac{dT}{T - T_a} = -k dt\)

Integrate both sides

Integrate the left side with respect to \(T\) and the right side with respect to \(t\): \(\int \frac{dT}{T - T_a} = \int -k dt\) This yields: \(\ln |T - T_a| = -kt + C\)

Solve for T(t)

Use the exponential function to solve for \(T(t)\): \(T(t) = T_a + (T_0 - T_a)e^{-kt}\) Now, let's find the temperature function based on the given data points: - \(T_0 = 100^{\circ} \mathrm{C}\) - \(T_a = 10^{\circ} \mathrm{C}\) - \(T(30) = 80^{\circ} \mathrm{C}\)

Determine k

Plug in the data points to find \(k\). Use the \(T(30)=80^{\circ} \mathrm{C}\) data point: \(80 = 10 + (100 - 10)e^{-30k}\) Solve for \(k\): \(70 = 90e^{-30k}\) \(\Rightarrow \frac{7}{9} = e^{-30k}\) \(\Rightarrow \ln\left(\frac{7}{9}\right) = -30k\) \(\Rightarrow k = \frac{\ln\left(\frac{9}{7}\right)}{30}\)

Find time when temperature reaches \(30^{\circ} \mathrm{C}\)

Use the found temperature function and solve for the time when the temperature reaches \(30^{\circ} \mathrm{C}\): \(30 = 10 + (100-10)e^{-\frac{t\ln(\frac{9}{7})}{30}}\) Solve for \(t\): \(\frac{20}{90} = e^{-\frac{t\ln(\frac{9}{7})}{30}}\) \(\Rightarrow \ln\left(\frac{2}{9}\right) = -\frac{t\ln(\frac{9}{7})}{30}\) \(\Rightarrow t = \frac{-30\ln\left(\frac{2}{9}\right)}{\ln(\frac{9}{7})}\) The time it takes for the soup to reach \(30^{\circ} \mathrm{C}\) is approximately: \(t \approx \frac{-30\ln\left(\frac{2}{9}\right)}{\ln(\frac{9}{7})}\) minutes.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are used to describe a wide range of phenomena, especially those related to rates of change. In the context of Newton's Law of Cooling, we deal with a specific type of differential equation that describes how the temperature of an object changes over time. This equation reflects how the rate of change in temperature is proportional to the difference between the object's temperature and the ambient temperature of its surroundings.

To solve such differential equations, especially the ones related to Newton's Law of Cooling, we often separate the variables and integrate. This involves splitting the equation into parts that can be independently processed for each variable and provides a solution to how the temperature changes over time. The goal is to find a function that models the temperature as a function of time, often leading to a derived exponential function.

In our exercise, we started by rearranging our differential equation using separation of variables, which is a common technique to simplify and solve these kinds of equations. This method allows us to isolate the rate of change and solve for the specific temperature function.

Temperature Function

The temperature function is an equation derived from solving the differential equation associated with Newton's Law of Cooling. It helps us predict the temperature of an object at any given time, based on initial conditions. The general form of the temperature function in this context is:

• \[ T(t) = T_a + (T_0 - T_a)e^{-kt} \]

Here, \(T(t)\) represents the temperature at time \(t\), \(T_a\) is the ambient temperature, \(T_0\) is the initial temperature of the object, and \(k\) is a constant that depends on the cooling properties of the object and its environment.

In our specific problem, we've been given some initial data points, such as the initial temperature of the soup and the temperature after 30 minutes. By plugging these values into our function, we can determine the cooling constant \(k\), which is crucial for predicting future temperatures.

Once we have the temperature function fully defined, it allows us to calculate when the soup will reach any desired temperature, such as the target 30 degrees Celsius. This ability to predict future states is a powerful application of differential equations in real-world scenarios.

Exponential Function

The exponential function plays a central role in modeling processes like cooling, where the change rate is proportional to the current state. This is because exponential functions can describe how quantities grow or decay at rates proportional to their size, which is a fundamental characteristic of many natural processes.

In the context of Newton's Law of Cooling, the exponential function describes how quickly or slowly the temperature of the object approaches the ambient temperature. This relationship is expressed mathematically in the temperature function, where the exponential term \(e^{-kt}\) indicates the influence of time on the cooling process.

The constant \(k\) in the exponential function directly affects the rate at which the temperature changes. A larger \(k\) value means a faster rate of cooling, while a smaller \(k\) indicates slower cooling. By adjusting \(k\), we can model the cooling for various conditions, making the exponential function highly adaptable and useful.

This function’s graph possesses a distinct curve that tends to flatten as time progresses, illustrating the deceleration in temperature change as it approaches equilibrium with the ambient environment. Understanding this behavior helps in forecasting not only cooling but also many other processes in physics and beyond.