Problem 17 - Advanced Engineering Mathematics

Question

A thermometer ceading \(70^{\circ} \mathrm{F}\) is placed in an oven preheated to a constant t-mperahire. Turough a glass window in the oven door, an observer records that the thermometer read \(110^{\circ} \mathrm{F}\) after \(\frac{1}{2}\) mimute and \(145^{\circ} \mathrm{F}\) after 1 minute. How hot is the oven?

Step-by-Step Solution

Answer

The oven is approximately \( 180^{\circ} \mathrm{F} \).

Step 1: Identify the Problem

We need to find the final constant temperature of the oven using Newton's Law of Cooling.

Step 1: Set Up the Newton's Law of Cooling Equation

Newton’s Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature (oven’s temperature in this case). The general formula is: \[ T(t) = T_s + (T_0 - T_s) e^{-kt} \] where \( T(t) \) is the temperature at time \( t \), \( T_s \) is the surrounding (oven) temperature, \( T_0 \) is the initial temperature, and \( k \) is the cooling constant.

Step 2: Plug in Known Values for Initial Reading

The thermometer starts at \( 70^{\circ} \mathrm{F} \) (\( T_0 \)); after 0.5 minutes, it's \( 110^{\circ} \mathrm{F} \) (\( T(0.5) = 110 \)). This gives us: \[ 110 = T_s + (70 - T_s) e^{-0.5k} \]

Step 3: Solve for Temperature Expression

Rearranging the first equation to solve for \( e^{-0.5k} \), we get: \[ 110 - T_s = (70 - T_s) e^{-0.5k} \] \[ e^{-0.5k} = \frac{110 - T_s}{70 - T_s} \]

Step 4: Plug in Second Reading

For \( T(1) = 145 \), we use: \[ 145 = T_s + (70 - T_s) e^{-k} \] Rearranging gives: \[ 145 - T_s = (70 - T_s) e^{-k} \] \[ e^{-k} = \frac{145 - T_s}{70 - T_s} \]

Step 5: Relate Both Equations

The relationship between \( e^{-0.5k} \) and \( e^{-k} \) leads to: \[ \left( \frac{110 - T_s}{70 - T_s} \right)^2 = \frac{145 - T_s}{70 - T_s} \]

Step 6: Solve for Oven Temperature \( T_s \)

By setting both expressions equal and solving for \( T_s \), we get: \[ \left(\frac{110 - T_s}{70 - T_s}\right)^2 = \frac{145 - T_s}{70 - T_s} \] Simplifying the equation will result in a quadratic formula, solving it provides \( T_s \).

Step 7: Calculation and Conclusion

Through solving the quadratic equation, we find that the surrounding temperature \( T_s \) is approximately \( 180^{\circ} \mathrm{F} \).

Key Concepts

Temperature Measurement

Measuring temperature is a fundamental aspect of science, providing insight into various processes and ensuring safety and efficiency in daily life. Thermometers are the primary tools for measuring temperature. They can be used in science labs, cooking, weather prediction, and medical practices.
Understanding how to properly measure temperature involves knowing how thermometers work. For example, a mercury thermometer functions on the principle of thermal expansion, where mercury expands and contracts with temperature changes.
In the exercise, temperature measurement is critical for understanding how quickly the thermometer can capture changes in the oven environment. Consistent temperature readings allow one to trace how quickly the thermometer's temperature moves toward the oven's constant temperature. This forms the basis for applying Newton's Law of Cooling in this scenario.

Differential Equations

Differential equations are a key mathematical tool used to describe how something changes over time. They consist of unknown functions and their derivatives, which make them ideal for modeling dynamic systems.
In the context of Newton's Law of Cooling, a differential equation helps describe how the temperature of the thermometer approaches the oven's constant temperature. The rate of temperature change depends on the difference between the object's temperature and the ambient temperature.
The law expressed as the equation: \[ T(t) = T_s + (T_0 - T_s) e^{-kt} \] illustrates how temperature changes over time, where \( T_s \) is the ambient temperature, \( T_0 \) is the initial temperature, and \( e^{-kt} \) represents the cooling effect over time, with \( k \) as a constant.

Quadratic Equations

Quadratic equations appear frequently in physics and engineering problems, typically taking the form \( ax^2 + bx + c = 0 \). These equations arise in this exercise when determining the constant temperature (\( T_s \)) in the oven.
The process of solving involves isolating \( T_s \) in our transformed relationship, which becomes a quadratic equation. This develops from the algebraic manipulation of terms when you equate the exponential expressions from the differential equation applications.
Solving these quadratic equations enables us to pinpoint the exact temperature of the oven, resulting from the given temperature measurements. By applying the quadratic formula, one can find the roots or solutions to these equations, revealing the constant temperature desired.

Proportionality Constant

The proportionality constant in the context of Newton's Law of Cooling is denoted by \( k \). This constant determines the rate at which the temperature change occurs, showing how quickly or slowly the thermometer adjusts to the ambient temperature.
In applications, \( k \) is crucial because it helps determine stability and speed in temperature changes. A higher \( k \) value means rapid adjustment, while a lower value indicates slower changes.
The exercise highlights the importance of this constant in ensuring that the thermometer readings match the expected rate of change based on the actual oven temperature. Solving for \( k \) provides insights into how efficiently the thermometer's temperature aligns with the expected behavior outlined by the law of cooling.