Problem 80 - Algebra and Trigonometry

Question

Cooling an Engine Suppose you're driving your car on a cold winter day \(\left(20^{\circ} \mathrm{F} \text { outside) and the engine overheats (at }\right.\) about \(220^{\circ} \mathrm{F} ) .\) When you park, the engine begins to cool down. The temperature \(T\) of the engine \(t\) minutes after you park satisfies the equation $$ \ln \left(\frac{T-20}{200}\right)=-0.11 t $$ (a) Solve the equation for \(T\) . (b) Use part (a) to find the temperature of the engine after 20 \(\min (t=20) .\)

Step-by-Step Solution

Answer

T(t) = 200e^{-0.11t} + 20; T(20 if t=20) ≈ 62.16°F.

Step 1: Isolate the Exponential Element

We start by isolating the logarithmic expression in the given equation. The equation given is \( \ln \left(\frac{T-20}{200}\right)=-0.11 t \). To isolate the exponential term, we use the property of logarithms that states that if \( \ln(a) = b \), then \( a = e^b \). This gives us: \[ \frac{T-20}{200} = e^{-0.11t} \].

Step 2: Solve for T

Next, we need to solve for \( T \). To do this, multiply both sides of the equation by 200:\[ T - 20 = 200 e^{-0.11t} \].Then, add 20 to both sides to isolate \( T \):\[ T = 200 e^{-0.11t} + 20 \].We now have an expression for \( T \):\[ T(t) = 200 e^{-0.11t} + 20 \].

Step 3: Calculate the Temperature after 20 Minutes

We now have a formula for \( T(t) \). To find the temperature of the engine after 20 minutes, substitute \( t = 20 \) into the expression for \( T(t) \):\[ T(20) = 200 e^{-0.11 \times 20} + 20 \].Calculate inside the exponent and the expression:\[ T(20) = 200 e^{-2.2} + 20 \].Using a calculator, find \( e^{-2.2} \) \( \approx 0.1108 \).Finally, compute:\[ T(20) = 200 \times 0.1108 + 20 \approx 42.16 + 20 = 62.16 \].Thus, the temperature of the engine after 20 minutes is approximately \( 62.16^{\circ} \text{F} \).

Key Concepts

Understanding Newton's Law of Cooling

Newton's Law of Cooling provides us with a mathematical framework for understanding how objects cool in an environment. This principle suggests that the rate at which an object cools is proportional to the difference in temperature between the object and the surrounding environment. In simple terms, the larger the temperature difference, the faster the object cools. This is why hot coffee cools faster initially than when it is lukewarm.

In the context of the given problem, the surrounding environment is the cold winter day temperature at 20°F. The car engine starts at a high temperature of 220°F. Newton's Law of Cooling helps us model the cooling process as a logarithmic equation, which in this case reflects the relationship between time and the engine's temperature as it returns towards the ambient temperature of 20°F. The cooling process is expressed through a logarithmic function that translates into an easily analyzed exponential function for solving practical cooling problems.

The Role of Exponential Functions

Exponential functions are a crucial element when modeling changes like cooling, where the rate of change is proportional to the current value. In our problem, the exponential function emerges from the logarithmic equation used to model the cooling of the engine. After isolating the logarithmic expression, we find that the exponential function we use is \[ T(t) = 200 e^{-0.11t} + 20 \].

This exponential function gives us the temperature of the engine, \( T(t) \), at any time \( t \). The base of the natural logarithm, \( e \), represents the continuous growth or decay rate. The negative exponent indicates a decay, illustrating how the temperature decreases over time. This represents an exponential decay because the rate of temperature change decreases as the temperature approaches the ambient temperature. This behavior is consistent with many real-world processes where exponential functions are applied to capture the change progressively lessens over time.

Approaches to Temperature Modeling

Modeling temperature changes over time is essential, not just in cars but in many heating and cooling applications. Temperature models need to accurately capture how temperatures evolve, possibly under varying environmental conditions. The model used in the exercise captures how rapidly or slowly temperature changes based on the initial temperature difference and environmental temperature.

When we use the model \( T(t) = 200 e^{-0.11t} + 20 \), it shows the expected temperature of the engine at any given time. Key aspects of this model include:

• The exponential decay component \( e^{-0.11t} \) that shows how the rate of cooling slows over time.
• The offset of 20°F which represents the starting temperature of the environment, influencing the asymptotic behavior of the function.

Using such models enables us to predict how long it will take for an object to reach a specific temperature or to understand the cooling dynamics under fixed conditions, providing valuable information for designing engines and predicting their thermal behavior.