Problem 46 - Advanced Engineering Mathematics

Question

Cooling and Warming A small metal bar is removed from an oven whose temperature is a constant \(300^{\circ} \mathrm{F}\) into a room whose temperature is a constant \(70^{\circ} \mathrm{F}\). Simultaneously, an identical metal bar is removed from the room and placed into the oven. Assume that time \(t\) is measured in minutes. Discuss: Why is there a future value of time, call it \(t^{*}>0\), at which the temperature of each bar is the same?

Step-by-Step Solution

Answer

There is a time \(t^{*}\) when both bars have the same temperature because they converge towards each other's initial temperatures under Newton's Law of Cooling.

Step 1: Understand the Problem

We have two identical metal bars, one initially taken from an oven at \(300^{\circ} \mathrm{F}\) and placed into a room at \(70^{\circ} \mathrm{F}\), and the other taken from the room and placed into the oven. We need to explore if there is a time \(t^{*}\) at which both bars reach the same temperature.

Step 2: Apply Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature. Let's denote the temperature of the bar from the oven as \(T_1(t)\) and the bar from the room as \(T_2(t)\). The room is at \(70^{\circ} \mathrm{F}\) and the oven is at \(300^{\circ} \mathrm{F}\).

Step 3: Setup Differential Equations

For the bar from the oven, we have \(\frac{dT_1}{dt} = -k(T_1 - 70)\) and for the bar from the room, \(\frac{dT_2}{dt} = -k(T_2 - 300)\), where \(k\) is a positive constant representing the cooling rate.

Step 4: Solve Differential Equations

Solving these equations, we use the formula for exponential decay in temperature. The solution for the temperature of the bar originally from the oven is \(T_1(t) = 70 + (300 - 70)e^{-kt}\). Similarly, for the bar originally in the room, \(T_2(t) = 300 - (300 - 70)e^{-kt}\).

Step 5: Find Time t* when Temperatures are Equal

Set \(T_1(t) = T_2(t)\) to find when the temperatures are the same: \(70 + 230e^{-kt} = 300 - 230e^{-kt}\). Solve for \(t\): \(230e^{-kt} + 230e^{-kt} = 230\). Therefore, \(460e^{-kt} = 230\) simplifies to \(e^{-kt} = \frac{1}{2}\). Taking the natural logarithm of both sides yields \(-kt = \ln(\frac{1}{2})\), giving \(t^* = -\frac{\ln(\frac{1}{2})}{k}\).

Step 6: Conclusion

There indeed exists a positive time \(t^{*}\) where the temperatures of the two bars are equal, which is understandable as both are changing temperature by the same rate constant and move toward each other's initial conditions.

Key Concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. In the context of Newton's Law of Cooling, these equations help us model how the temperature of an object changes over time as it exchanges heat with its environment. The key to using differential equations is understanding that they express the rate at which a situation changes.
In the problem of two metal bars exchanged between an oven and a room, we set up differential equations to capture how their temperatures adjust. For the metal bar from the oven, the differential equation \[ \frac{dT_1}{dt} = -k(T_1 - 70) \] describes how its temperature decreases towards the room temperature of \(70^{\circ} \mathrm{F}\). Similarly, for the bar from the room, \[ \frac{dT_2}{dt} = -k(T_2 - 300) \] shows how it increases towards the oven temperature of \(300^{\circ} \mathrm{F}\).
These equations are dynamic, meaning they evolve with time, and the rate \(-k\) describes how quickly these changes occur, depending on how different the current temperature is from the surroundings. To find a tangible solution (i.e., exact temperatures at different times), we solve these equations.

Exponential Decay

Exponential decay is a process where the value of something decreases at a rate proportional to its current value. It is common in processes like cooling, radioactive decay, and interest reduction. In the context of Newton's Law of Cooling, exponential decay describes how temperatures adjust over time to reach a stable state.
The formula outputs from the differential equations, \[ T_1(t) = 70 + (300 - 70)e^{-kt} \] and \[ T_2(t) = 300 - (300 - 70)e^{-kt} \], reflect exponential decay behavior. Here, the expressions \(e^{-kt}\) show how fast the temperature changes occur, slowing down as time progresses. The constant \(k\), again, defines the rate of temperature change.
This pattern explains why the temperatures never stay constant but gradually adjust until they approach the temperature equilibrium, where the rates of increase and decrease become negligible.

Temperature Equilibrium

Temperature equilibrium occurs when two substances reach the same temperature and therefore no longer exchange heat. It is a state of balance. In the scenario of the two metal bars, a time \(t^{*}\) is found when both bars have equal temperatures, meaning they are in thermal equilibrium.
To find this moment, we set the solutions of the differential equations equal: \[ 70 + 230e^{-kt} = 300 - 230e^{-kt} \]. Simplifying this equation leads us to \(e^{-kt} = \frac{1}{2}\), which reveals \(t^{*} = -\frac{\ln(\frac{1}{2})}{k}\).
The concept of temperature equilibrium in this experiment becomes clear, as this specific time \(t^{*}\) indicates when the heat gained and lost by each bar balances exactly. This dynamic reflection of equilibrium shows the natural tendency of systems to stabilize over time.