Problem 23 - Thomas Calculus

Question

Surrounding medium of unknown temperature \(\mathrm{A}\) pan of warm water \(\left(46^{\circ} \mathrm{C}\right)\) was put in a refrigerator. Ten minutes later, the water's temperature was \(39^{\circ} \mathrm{C} ; 10 \mathrm{min}\) after that, it was \(33^{\circ} \mathrm{C}\) . Use Newton's law of cooling to estimate how cold the refrigerator was.

Step-by-Step Solution

Answer

The temperature of the refrigerator is approximately \(20^{\circ} \mathrm{C}\).

Step 1: Define 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. This can be mathematically expressed as \( \frac{dT}{dt} = -k (T - T_m) \), where \( T \) is the temperature of the object, \( T_m \) is the ambient temperature, and \( k \) is a constant.

Step 2: Write Down the Differential Equation

Given the change in temperature from \(46^{\circ} \mathrm{C}\) to \(39^{\circ} \mathrm{C}\) in 10 minutes, and then from \(39^{\circ} \mathrm{C}\) to \(33^{\circ} \mathrm{C}\) in another 10 minutes, we first need to set up the equation \( \frac{dT}{dt} = -k(T - T_m) \) for each interval.

Step 3: Solve for the Constant \(k\)

Start with the first interval. The temperature decays from \(46^{\circ} \mathrm{C}\) to \(39^{\circ} \mathrm{C}\) in 10 minutes. Using the formula \( T(t) = T_m + (T_0 - T_m)e^{-kt} \), where \( T_0 = 46^{\circ} \mathrm{C} \), and \( T(10) = 39^{\circ} \mathrm{C} \), solve for \(k\). Similarly, set up the equation for the second interval using \( T(20) = 33^{\circ} \mathrm{C} \) and solve for \(k\).

Step 4: Establish Equations for Both Intervals

For the first 10 minutes:\[ 39 = T_m + (46 - T_m)e^{-10k} \]For the second 10 minutes:\[ 33 = T_m + (39 - T_m)e^{-10k} \]

Step 5: Solve the System of Equations

With the equations from Step 4, solve the system of simultaneous equations to find the value of \( T_m \). This provides us with the ambient temperature that the water is approaching, which represents the refrigerator's temperature.

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates a function with its derivatives. In Newton's Law of Cooling, the differential equation expresses how the temperature of an object changes over time in relation to the surrounding environment. This relationship is significant because it allows us to predict temperature changes.
In this context, the differential equation is represented as:

• \( \frac{dT}{dt} = -k (T - T_m) \)

Here, \( T \) is the temperature of the object at time \( t \), \( T_m \) is the ambient temperature, and \( k \) is a constant that depends on the characteristics of the object and the environment.

By solving this differential equation, we can find how quickly an object cools down or heats up, depending on the temperature of the surroundings.

Ambient Temperature

The ambient temperature, denoted by \( T_m \) in Newton's Law of Cooling, is the constant temperature of the surrounding environment towards which an object's temperature moves over time. Understanding ambient temperature is crucial in applying the law effectively.

It acts as the baseline that dictates the direction of heat transfer. If the ambient temperature is cooler than the object, as in the given exercise, the object will lose heat.

• The object will eventually approach the ambient temperature but practically never reach it due to the nature of exponential decay.

The problem of estimating how cold the refrigerator was revolves around finding this ambient temperature value using the given cooling data.

Exponential Decay

Exponential decay is a process where quantities decrease at a rate proportional to their current value. This is a fundamental concept in Newton's Law of Cooling, where the rate of temperature change of an object is proportional to the difference between the object's temperature and the ambient temperature.

In our exercise, the formula used is:

• \( T(t) = T_m + (T_0 - T_m)e^{-kt} \)

This equation describes how the temperature \( T \) decreases exponentially over time \( t \) toward the ambient temperature \( T_m \).

The term \( e^{-kt} \) signifies the exponential decay factor, and \( k \) is the decay constant. This constant determines the rate at which the temperature approaches the ambient temperature. A larger \( k \) means the temperature drops faster. Understanding exponential decay in this context helps predict how rapidly an object's temperature will change when placed in a different environment.