Problem 27 - 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. A body is discovered in a hotel room. At 7: 00 A.M. a police detective found the body's temperature to be \(85^{\circ} \mathrm{F}\). At 8: 30 A.M. a medical examiner measures the body's temperature to be \(82^{\circ} \mathrm{F}\). Assuming the room in which the body was found had a constant temperature of \(74^{\circ} \mathrm{F}\), how long has the victim been dead? (Normal body temperature is \(98.6^{\circ} \mathrm{F} .\) ).

Step-by-Step Solution

Answer

The victim died around 5:55 A.M.

Step 1: Identify Known Values

We need to identify all known values from the problem.- At 7:00 A.M. (we can take this as \( t = 0 \)), the body temperature \( T(0) = 85^{\circ} \mathrm{F} \).- At 8:30 A.M., \( T(1.5) = 82^{\circ} \mathrm{F} \).- The room temperature, \( T_{S} = 74^{\circ} \mathrm{F} \).- The initial body temperature when the victim died, \( T_{0} = 98.6^{\circ} \mathrm{F} \).

Step 2: Apply Newton’s Law of Cooling Formula

Using the cooling formula: \[ T = T_{S} + (T_{0} - T_{S}) e^{-k t} \] Insert known values to create two equations:1. At \( t = 0 \): \[ 85 = 74 + (98.6 - 74) e^{-k \cdot 0} \] Which simplifies to: \[ 85 = 74 + 24.6 \] This confirms the initial equation is satisfied at \( t = 0 \).2. At \( t = 1.5 \): \[ 82 = 74 + (98.6 - 74) e^{-1.5k} \]

Step 3: Solve for k

Use the equation \[ 82 = 74 + 24.6 e^{-1.5k} \]:- Subtract 74 from both sides: \[ 8 = 24.6 e^{-1.5k} \]- Divide both sides by 24.6: \[ \frac{8}{24.6} = e^{-1.5k} \]- Take the natural logarithm of both sides to solve for \( k \): \[ \ln\left(\frac{8}{24.6}\right) = -1.5k \]- Calculate the numerical values: \[ \ln(0.3252) = -1.5k \] \[ -1.122 = -1.5k \]- Solve for \( k \): \[ k \approx 0.748 \]

Step 4: Determine Time of Death

Now use the value of \( k \) to find how long the body has been cooling. Since normal body temperature is \( 98.6^{\circ} \mathrm{F} \), we want to solve the equation for \( t \) when \( T(t) = 98.6 \). However, that is the initial condition before cooling. Therefore, initially the temperature was decreasing from \( 98.6 \).Use \[ 85 = 74 + (98.6 - 74) e^{-kt} \]:- Simplify: \[ 85 = 74 + 24.6 e^{-0.748t} \]- Rearrange the equation: \[ 11 = 24.6 e^{-0.748t} \]- Divide by 24.6: \[ \frac{11}{24.6} = e^{-0.748t} \]- Take the natural logarithm of both sides: \[ \ln\left(\frac{11}{24.6}\right) = -0.748t \]- Calculate the log and solve for \( t \): \[ \ln(0.447) = -0.748t \] \[ -0.805 = -0.748t \]- Solve for \( t \): \[ t \approx 1.08 \text{ hours before 7:00 A.M.} \]

Step 5: Calculate Exact Death Time

Since \( t \approx 1.08 \) hours before 7:00 A.M., we calculate the exact time:1.08 hours translates to approximately 1 hour and 5 minutes.Therefore, the time of death is approximately 5:55 A.M.

Key Concepts

Temperature Change Modeling

Newton's Law of Cooling is a fantastic example of temperature change modeling in real-world scenarios. When you have an object that starts with a temperature different from its surroundings, it will eventually change to match the ambient temperature. This process is what we model using mathematical equations.

The basic principle is straightforward: the rate at which an object’s temperature changes is proportional to the difference between its current temperature and the environment’s temperature. This principle allows us to predict how quickly, for example, a hot cup of coffee will cool down in a room.

By applying Newton's Law of Cooling, we can set up a model that predicts temperature changes over time. The formula uses several variables:

• Time ( t )
• Initial object temperature ( T_{0} )
• Ambient temperature ( T_{S} )
• A rate constant ( k )

This model provides the exponential function that fits the change in temperature over time, which is very powerful when you have initial temperature values and need to know future temperatures.

Exponential Decay

Exponential decay is a term you'll encounter frequently with processes that reduce in value over time. When we talk about temperature dropping, we're often describing an exponential decay process.

In Newton's Law of Cooling, the temperature change follows the formula \( T = T_{S} + (T_{0} - T_{S}) e^{-k t} \), where \( e^{-k t} \) represents the decay of temperature over time. Essentially, as time ( t ) increases, the exponential decay factor causes the temperature to approach the surrounding temperature.

Exponential decay is characterized by:

• A steady decrease by a constant percentage over equal time intervals.
• An initial value that decreases more rapidly at first before slowing as it approaches the equilibrium.

Understanding this helps us recognize why a hot object cools slower as it approaches ambient temperature and why it's crucial for calculations related to temperature changes.

Temperature Equilibria

In any temperature change scenario where Newton’s Law of Cooling is applied, it's essential to consider temperature equilibria. This term refers to the point at which the temperature of an object matches the ambient temperature, and cooling or heating ceases unless external factors change.

In our example, the body or object cools until it reaches the room temperature of \( 74^{\circ} \mathrm{F} \). That is the equilibrium state, where energy exchange between the body and room stops because there is no longer a temperature gradient driving the change.

Key points about temperature equilibria:

• It marks the balance point between the object and its environment.
• This state achieves a dynamic equilibrium, meaning temperatures can fluctuate but stay close to this equilibrium point unless affected by outside forces.
• Understanding equilibria is vital for determining practical endpoints in temperature-related calculations, such as estimating time of death in forensic investigations based on cooling bodies.

The concept of equilibrium helps set tangible limits to the application of Newton’s Law in diverse situations, making it a reliable tool in scientific calculations.