Question

At 10 A.M. a woman took a cup of hot instant coffee from her microwave oven and placed it on a nearby kitchen counter to cool. At this instant the temperature of the coffee was \(180^{\circ} \mathrm{F}\), and 10 minutes later it was \(160^{\circ} \mathrm{F}\). Assume the constant temperature of the kitchen was \(70^{\circ} \mathrm{F}\). (a) What was the temperature of the coffee at \(10: 15\) A.m.? (b) The woman of this problem likes to drink coffee when its temperature is between \(130^{\circ} \mathrm{F}\) and \(140^{\circ} \mathrm{F}\). Between what times should she have drunk the coffee of this problem?

Short answer

At 10:15 A.M., the temperature of the coffee was approximately \(145.65^{\circ} \mathrm{F}\). The woman should drink her coffee when the time is between \(11.45\) minutes (approximately 10:11 A.M.) and \(16.87\) minutes (approximately 10:17 A.M.) after taking it from the microwave.

Step-by-step solution

Find the Initial Conditions and Constants

Here, we know the following information: Initial Temperature of coffee, \(T_0 = 180^{\circ} \mathrm{F}\) Temperature of coffee after 10 minutes, \(T_{10} = 160^{\circ} \mathrm{F}\) Temperature of kitchen, \(T_m = 70^{\circ} \mathrm{F}\) Newton's Law of Cooling is given by the formula: \[ \frac{dT}{dt} = -k(T - T_m) \] Where \(T\) is the temperature of the coffee, \(t\) is time in minutes, and \(k\) is a proportionality constant. We need to find the values of \(k\) and the equation to find the temperature for any given time.

Solve the Differential Equation

To find the equation to calculate the temperature at any time, we will solve the differential equation from Newton's Law of Cooling. \[ \frac{dT}{dt} = -k(T - T_m) \] First, rearrange the equation to separate the variables: \[ \frac{dT}{T - T_m} = -k dt \] Now, integrate both sides of the equation: \[ \int \frac{dT}{T - T_m} = -\int k dt \] \[ \ln|(T - T_m)| = -kt + C \] Next, exponentiate both sides: \[ T - T_m = Ke^{-kt} \] We have now found a general solution for the equation. Now we need to find the specific solution by finding the value of \(K\) using the initial condition.

Find the Specific Solution

Using the initial condition, \(T_0 = 180^{\circ} \mathrm{F}\), when \(t = 0\), we can find the value of \(K\): \[ 180 - 70 = Ke^{-k\cdot 0} \] \[ 110 = K \] The specific solution of the temperature equation is: \[ T(t) = 110e^{-kt} + 70 \] Now we need to find the value of \(k\) using the condition \(T_{10} = 160^{\circ} \mathrm{F}\), when \(t = 10\).

Find the Value of k

Using the condition \(T_{10} = 160\), when \(t = 10\): \[ 160 = 110e^{-10k} + 70 \] Now, solve for \(k\): \[ 90 = 110e^{-10k} \] \[ \frac{9}{11} = e^{-10k} \] \[ \ln\left(\frac{9}{11}\right) = -10k \] \[ k = \frac{-\ln\left(\frac{9}{11}\right)}{10} \] Now we have the value of \(k\) and can write the full equation for the temperature of the coffee as a function of time.

Temperature at 10:15 A.M. (a)

We have the equation for the temperature of the coffee as a function of time: \[ T(t) = 110e^{-\frac{-\ln\left(\frac{9}{11}\right)}{10}t} + 70 \] When \(t = 15\) minutes, we can find the temperature at 10:15 A.M: \[ T(15) = 110e^{-\frac{-\ln\left(\frac{9}{11}\right)}{10}\cdot 15} + 70 \approx 145.65^{\circ} \mathrm{F} \] At 10:15 A.M., the temperature of the coffee was approximately \(145.65^{\circ} \mathrm{F}\).

Time to Drink Coffee (b)

The woman likes to drink coffee when its temperature is between \(130^{\circ} \mathrm{F}\) and \(140^{\circ} \mathrm{F}\). We will use the temperature equation to find the time at which the coffee will achieve these temperatures. When \(T = 130\): \[ 130 = 110e^{-\frac{-\ln\left(\frac{9}{11}\right)}{10}t} + 70 \] Solve for \(t\): \[ 60 = 110e^{-\frac{-\ln\left(\frac{9}{11}\right)}{10}t} \] \[ \frac{6}{11} = e^{-\frac{-\ln\left(\frac{9}{11}\right)}{10}t} \] \[ t_1 = \frac{-10\ln\left(\frac{6}{11}\right)}{\ln\left(\frac{9}{11}\right)} \approx 16.87 \text{ minutes} \] When \(T = 140\): \[ 140 = 110e^{-\frac{-\ln\left(\frac{9}{11}\right)}{10}t} + 70 \] Solve for \(t\): \[ 70 = 110e^{-\frac{-\ln\left(\frac{9}{11}\right)}{10}t} \] \[ \frac{7}{11} = e^{-\frac{-\ln\left(\frac{9}{11}\right)}{10}t} \] \[ t_2 = \frac{-10\ln\left(\frac{7}{11}\right)}{\ln\left(\frac{9}{11}\right)} \approx 11.45 \text{ minutes} \] The woman should drink her coffee when the time is between \(11.45\) minutes (approximately 10:11 A.M.) and \(16.87\) minutes (approximately 10:17 A.M.) after taking it from the microwave.

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations that involve functions and their derivatives. In a physical context, they describe how a certain quantity changes over time or space. Newton's Law of Cooling, as seen in the exercise, uses an ODE to model the change in temperature of an object over time.

Specifically, the law states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature, assuming no other heat transfer. This is represented by the ODE \[\frac{dT}{dt} = -k(T - T_m)\]where \(T\) is the variable temperature, \(T_m\) is the ambient temperature, \(t\) is time, and \(k\) is a positive constant characteristic of the object. The negative sign indicates that the temperature decreases over time if the object is hotter than the surroundings. ODEs are significant in modeling many natural phenomena and require specific methods for their solutions.

Solving Differential Equations

Solving ordinary differential equations (ODEs) often involves finding a function that satisfies the given relationship between its various derivatives. The process typically requires integration, separation of variables, applying initial conditions, and sometimes more advanced techniques such as Laplace transforms or series expansions, depending on the complexity of the equation.

In the context of Newton's Law of Cooling, the differential equation is first order and separable. By organizing all the \(T\) terms on one side and the \(t\) terms on the other, we can integrate each side separately, which leads to a logarithmic function. Exponentiating both sides then gives us a general solution, which involves an arbitrary constant \(K\).

The initial condition is then used to find the value of \(K\), and any additional conditions can help determine the value of other parameters, like the constant \(k\) in our exercise. This part of the process showcases how additional information about the system behavior at specific points in time is crucial for specifying a unique solution to an ODE.

Exponential Decay

Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value, resulting in a rapid decline that gradually slows over time. This concept is crucial in various fields, such as physics, chemistry, and finance. The general formula for exponential decay is \[N(t) = N_0e^{-kt}\]where \(N(t)\) is the quantity at time \(t\), \(N_0\) is the initial quantity, \(k\) is the decay constant, and \(e\) is the base of natural logarithms.

In our exercise involving Newton's Law of Cooling, the temperature of the coffee exhibits exponential decay as it approaches the ambient temperature of the room. The solution provided demonstrates calculating the temperature at a particular time as well as determining when the temperature falls within a desirable range for drinking, both applications of exponential decay. It illustrates how fundamentally exponential decay, described by ODEs, is at predicting time-based behaviors in natural processes.