Problem 40 - Calculus and its applications

Question

After warming the water in a hot tub to \(100^{\circ},\) the heating element fails. The surrounding air temperature is \(40^{\circ},\) and in 5 min the water temperature drops to \(95^{\circ}\) a) Find the value of the constant \(a\) in Newton's Law of Cooling. b) Find the value of the constant \(k\). Round to five decimal places. c) What is the water temperature after \(10 \mathrm{~min} ?\) d) How long does it take the water to cool to \(41^{\circ} ?\) e) Find the rate of change of the water temperature, and interpret its meaning.

Step-by-Step Solution

Answer

a) 60, b) 0.01704, c) ~90.42°C, d) ~168.38 min, e) -1.0224°C/min: temperature decreases initially by ~1.0224°C/min.

Step 1: Understanding Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between the object's temperature and the ambient temperature. Mathematically, it can be expressed as \( \frac{dT}{dt} = -k(T - T_s) \) where \(T\) is the temperature of the object, \(T_s\) is the surrounding temperature, \(k\) is a constant, and \(t\) is time. We will use this formula to solve the subsequent parts.

Step 2: Formulating the Temperature Equation

The solution to the differential equation from Newton's Law is \( T(t) = T_s + (T_0 - T_s)e^{-kt} \), where \(T_0\) is the initial temperature. In this problem, \(T_0 = 100^{\circ}\) and \(T_s = 40^{\circ}\). Substitute these values to get the equation: \( T(t) = 40 + (100 - 40)e^{-kt} \).

Step 3: Finding the Constant a

The constant \(a\) here refers to the difference between initial water temperature and the ambient temperature, which is the initial difference driving the heat exchange. Therefore, \(a = T_0 - T_s = 100 - 40 = 60\).

Step 4: Finding the Constant k

Use the given condition that after 5 minutes the temperature dropped to 95°C to find \(k\). Substitute into the temperature equation: \( 95 = 40 + 60e^{-5k} \). Solve for \(k\):\[ 55 = 60e^{-5k} \]\[ e^{-5k} = \frac{55}{60} \]\[ -5k = \ln\left(\frac{55}{60}\right) \]\[ k = -\frac{1}{5}\ln\left(\frac{55}{60}\right) \approx 0.01704 \] (rounded to five decimal places).

Step 5: Calculating Water Temperature after 10 Minutes

Use the temperature equation with \(k = 0.01704\) to find \(T(10)\):\[ T(10) = 40 + 60e^{-0.1704} \]Calculate:\[ T(10) = 40 + 60 \times e^{-0.1704} \approx 90.42^{\circ} \] (after rounding).

Step 6: Determining Time to Cool to 41°C

Set \(T(t) = 41\) and solve for time \(t\):\[ 41 = 40 + 60e^{-kt} \]\[ 1 = 60e^{-kt} \]\[ e^{-kt} = \frac{1}{60} \]\[ -kt = \ln\left(\frac{1}{60}\right) \]\[ t = -\frac{1}{k}\ln\left(\frac{1}{60}\right) \approx \frac{1}{0.01704}\ln(60) \approx 168.38 \] (rounding to two decimal places, in minutes).

Step 7: Finding the Rate of Change of the Water Temperature

The rate of change of temperature is given by \( \frac{dT}{dt} = -k(T - T_s) \). Substitute \(T = 100 \) to find the initial rate:\[ \frac{dT}{dt} = -0.01704(100 - 40)\]\[ \frac{dT}{dt} = -1.0224 \]This indicates the water temperature changes by approximately \(-1.0224^{\circ}/\text{min}\) initially, meaning the temperature decreases by about \(1.0224^{\circ}\) per minute when it starts cooling.

Key Concepts

Differential Equations

Differential equations are a powerful tool in mathematics used to describe how quantities change over time. In the case of Newton's Law of Cooling, they help us understand how the temperature of a substance changes as it approaches equilibrium with its surroundings. The general form we use in cooling problems is \( \frac{dT}{dt} = -k(T - T_s) \), where \( T \) is the temperature of the object, \( T_s \) is the ambient temperature, and \( k \) is a constant that describes how quickly the cooling occurs.
To solve these differential equations, we integrate them over time to find an expression for temperature as a function of time, \( T(t) \). This process helps us predict how the temperature will change in future time intervals.
Differential equations can be complex, but in cooling problems, they provide a smooth way to model real-world decay in temperature with great accuracy.

Cooling Rate

The cooling rate refers to how quickly an object's temperature decreases towards the ambient temperature. According to Newton's Law of Cooling, this rate is directly proportional to the temperature difference between the object and its environment.
To quantify this, we use the cooling constant \( k \). It's a specific value that signifies the speed of cooling under given conditions. A greater \( k \) means the object will cool faster. In mathematical terms, the cooling rate at any point is captured by the expression \( \frac{dT}{dt} = -k(T - T_s) \).

• The present temperature of the object affects the rate, emphasizing that cooling occurs more rapidly when there is a large temperature gradient.
• The value of \( k \) can be derived from experimental data, such as measuring temperature changes at set intervals.
• An object will cool slower as the temperature approaches the ambient level, illustrating the diminishing return effect in natural cooling processes.

The cooling rate offers insight into thermal behaviors, essential for applications in both natural and industrial settings.

Temperature Change

Understanding how temperature changes over time is integral to solving problems using Newton's Law of Cooling. The temperature change of an object in a cooler environment can be expressed through the equation:

\[T(t) = T_s + (T_0 - T_s)e^{-kt}\]
Here, \( T_0 \) is the initial temperature, and \( T_s \) is the ambient temperature. The term \((T_0 - T_s)e^{-kt}\) describes the changing difference between the object's temperature and the ambient temperature.
The exponential decay function \(e^{-kt}\) models the rate at which this temperature difference closes over time. As the value of the expression decreases, the object's temperature comes closer to its surroundings. This continues until thermal equilibrium is achieved.
An important aspect of analyzing temperature change is predicting future temperatures, which helps in setting expectations in thermal management scenarios. The exponential nature of such changes highlights how adjustments happen quickly at first, diminishing progressively as the two temperatures converge.

Newton's Law

Newton's Law of Cooling is a principle stating that an object's temperature changes at a rate proportional to the difference between its current temperature and the ambient temperature. This law is expressed through the equation \( \frac{dT}{dt} = -k(T - T_s) \).

• The law assumes that the rate of heat loss from an object is dependent purely on this difference, making it a straightforward yet powerful tool for analyzing heat transfer.
• Newton's Law of Cooling is applicable when the ambient temperature remains constant over time and the object has a uniform temperature throughout.
• This law is used in a variety of scientific and engineering fields such as meteorology, culinary sciences, and even forensic investigations, to estimate time of death based on body cooling.

Understanding Newton's Law allows us to model real-world situations involving heat exchange effectively. Despite its simplicity, it accurately captures the essence of natural cooling processes, offering valuable predictions in practical situations.