Problem 28 - Precalculus

Question

Use the following formula for Newton’s Law of Cooling: If you take a hot dinner out of the oven and place it on the kitchen countertop, the dinner cools until it reaches the temperature of the kitchen. Likewise, a glass of ice set on a table in a room eventually melts into a glass of water at that room temperature. The rate at which the hot dinner cools or the ice in the glass melts at any given time is proportional to the difference between its temperature and the temperature of its surroundings (in this case, the room). This is called Newton's law of cooling (or warming) and is modeled by $$T=T_{S}+\left(T_{0}-T_{S}\right) e^{-k t}$$ where \(T\) is the temperature of an object at time \(t, T_{s}\) is the temperature of the surrounding medium, \(T_{0}\) is the temperature of the object at time \(t=0, t\) is the time, and \(k\) is a constant. At 4 A.M. a body is found in a park. The police measure the body's temperature to be \(90^{\circ} \mathrm{F} .\) At 5 A.M. the medical examiner arrives and determines the temperature to be \(86^{\circ} \mathrm{F}\). Assuming the temperature of the park was constant at \(60^{\circ} \mathrm{F}\), how long has the victim been dead?

Step-by-Step Solution

Answer

The victim died approximately at 1:47 AM.

Step 1: Identify Known Values

We are given the initial temperature of the body at 4 AM, which is \( T_0 = 90^{\circ}F \). At 5 AM, the temperature \( T = 86^{\circ}F \). The temperature of the surroundings (the park) is constant, \( T_s = 60^{\circ}F \).

Step 2: Equation at 5 AM

Using the formula \( T = T_{s} + (T_0 - T_{s})e^{-kt} \) and substituting \( T = 86 \), \( T_s = 60 \), and \( T_0 = 90 \), we have: \[ 86 = 60 + (90 - 60)e^{-k(1)} \] This simplifies to: \[ 26 = 30e^{-k} \]

Step 3: Solve for Constant k

We need to solve for \( e^{-k} \). Dividing both sides by 30, we get \[ e^{-k} = \frac{26}{30} = \frac{13}{15} \] Taking the natural logarithm on both sides: \[ -k = \ln\left(\frac{13}{15}\right) \] \[ k = -\ln\left(\frac{13}{15}\right) \approx 0.143 \]

Step 4: Equation at 4 AM

Given that the initial measurement at 4 AM was \( T_0 = 90 \) when \( t = 0 \), we now need to find when the body was at approximately \( 98.6^{\circ}F \), which is normal body temperature. So setting up: \[ 90 = 60 + (98.6 - 60)e^{-kt} \]

Step 5: Solve for Time of Death

Rearrange to \[ 30 = 38.6e^{-kt} \] Divide both sides by 38.6 to get \[ e^{-kt} = \frac{30}{38.6} \] Taking the natural logarithm: \[ -kt = \ln\left(\frac{30}{38.6}\right) \] Using \( k \approx 0.143 \), solve for \( t \): \[ t = \frac{-\ln\left(\frac{30}{38.6}\right)}{0.143} \approx 2.22 \text{ hours} \]

Step 6: Calculate Time of Death

Since \( t \approx 2.22 \) hours before 4 AM, the time of death is approximately 1:47 AM.

Key Concepts

Temperature Gradient

The idea behind Newton's Law of Cooling is heavily reliant on the concept of the temperature gradient. A temperature gradient is the difference in temperature between two points, which in our case is the body and the surroundings.
In Newton's Law of Cooling, the rate at which an object's temperature changes depends on this gradient.

• When you place a hot object, like a freshly baked pie, on a kitchen counter, it quickly releases heat to its cooler surrounding.
• The opposite happens with a cold object, such as a glass of ice, which absorbs heat from the room, as the temperature gradient exists there as well.

With the cooling body scenario in the park, the rate at which the body's temperature decreases is proportional to how much higher it is compared to the park's constant temperature of 60°F. As the temperatures become more similar, the gradient diminishes, slowing the cooling process as it approaches thermal equilibrium.

Exponential Decay

Exponential decay is the mathematical concept that describes how quantities decrease at a rate proportional to their current amount. In the context of cooling, as time progresses, the change in temperature of an object decreases exponentially.Newton's Law of Cooling includes the term \(e^{-kt}\), where \(k\) is a constant that determines how rapidly the cooling or warming occurs.

• When \(t = 0\), right at the start, the object is the furthest from reaching its surrounding temperature, indicating a steep descent in heat.
• As time goes on, \(e^{-kt}\) declines exponentially, decreasing the rate of temperature change.

In our case, \(k\) is found using logs, which illustrates that the decrease in body temperature from 90°F to 86°F is an instance of exponential decay, showing how quickly the body cools towards the ambient temperature of the park.

Thermal Equilibrium

Thermal equilibrium occurs when two objects reach the same temperature, and thus, there is no net heat flow between them. This is the final state that Newton's Law of Cooling predicts. For an object that is warmer or cooler than its environment, like a body cooling in the park, it will continue to change its temperature until it matches that of the surroundings.

• The body will continue to cool until it reaches the park's steady temperature of 60°F.
• Reaching thermal equilibrium means that the temperature gradient has been entirely diminished.

In real-life scenarios, reaching absolute thermal equilibrium is more theoretical, as slight fluctuations can occur, but objects will effectively stop cooling or warming once they closely match the temperature of their environment.