Question

You have a cup of coffee at temperature ${70}^{\circ }\mathrm{C}$ and the ambient temperature in the room is ${20}^{\circ }\mathrm{C}$. Assuming a cooling rate $k$ of $0.125$, write and solve the differential equation to describe the temperature of the coffee with respect to time.

Short answer

The temperature of the coffee as a function of time is $$T(t)=50e^{-0.125t}+20.$$

Step-by-step solution

Identify the Differential Equation as Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of an object's temperature is proportional to the difference between its own temperature and the ambient temperature. Mathematically, it is represented as:$$\frac{dT}{dt}=-k(T-T_{ambient})$$where $T$ is the temperature of the object (coffee), $T_{ambient}$ is the ambient temperature, and $k$ is the cooling constant.

Substitute Known Values

Given that the initial temperature of the coffee is ${70}^{\circ }C$, the ambient temperature is ${20}^{\circ }C$, and $k=0.125$. Substitute these values into the differential equation:$$\frac{dT}{dt}=-0.125(T-20)$$

Separate Variables

To solve this differential equation, separate the variables by moving all terms involving $T$ to one side of the equation and $dt$ to the other:$$\frac{dT}{T-20}=-0.125dt$$

Integrate Both Sides

Integrate both sides of the equation to solve for $T$:$$\int \frac{1}{T-20}dT=-0.125\int dt$$This results in:$$\ln |T-20|=-0.125t+C$$where $C$ is the constant of integration.

Solve for the Constant of Integration

Use the initial condition $T(0)=70$ to find $C$. Substitute $T=70$ and $t=0$ into the integrated equation:$$\ln |70-20|=-0.125(0)+C\Rightarrow C=\ln (50)$$

Solve for Temperature as a Function of Time

Substitute $C$ back into the equation:$$\ln |T-20|=-0.125t+\ln (50)$$Exponentiate both sides to solve for $T$:$$|T-20|=e^{-0.125t}\cdot 50$$This gives:$$T=50e^{-0.125t}+20$$ after considering the positive solution since temperature cannot be negative.

Key concepts

Understanding Differential Equations in Newton's Law of Cooling

Differential equations are fundamental tools for modeling phenomena where rate of change is involved. In the context of Newton's Law of Cooling, we are dealing with how the temperature of an object, like a cup of coffee, changes over time. The differential equation for this phenomenon is expressed as:$$\frac{dT}{dt}=-k(T-T_{ambient})$$Here, $\frac{dT}{dt}$ represents the rate of change of the object's temperature $T$ with respect to time $t$. The term $T_{ambient}$ is the constant ambient temperature, and $k$ is a positive constant that characterizes the cooling rate. This equation tells us that the rate at which the coffee cools is directly proportional to the difference in temperature between the coffee and its surrounding environment.

• When $T>T_{ambient}$, the coffee cools down (the rate of change $\frac{dT}{dt}$ is negative).
• When $T=T_{ambient}$, the temperature stabilizes ($\frac{dT}{dt}=0$).
• When $T<T_{ambient}$, which isn't typically relevant in this context, the coffee would warm up.

Temperature Change in Newton's Cooling Process

The temperature change in Newton's Law of Cooling is an exciting concept. It explains how heat dissipates from an object to its surroundings. In our exercise, the initial temperature of the coffee is ${70}^{\circ }C$, while the ambient room temperature is maintained at ${20}^{\circ }C$. This means that:$$T_{initial}={70}^{\circ }C\text{and}{T}_{ambient}={20}^{\circ }C$$As the coffee cools over time, its temperature difference with the ambient temperature decreases, following an exponential decay trend. The constant $k=0.125$ in our exercise plays a significant role in defining how quickly the temperature of the coffee drops.

• The larger the difference between $T$ and $T_{ambient}$, the faster the cooling process.
• As $T$ approaches $T_{ambient}$, the rate of temperature change slows down.
• Ultimately, the coffee's temperature will stabilize at the ambient room temperature.

By understanding this relationship, you can predict how long it will take for any hot beverage to reach a certain temperature.

Integrating to Solve for Temperature Function Over Time

Integration is a key mathematical process used to solve the differential equation. When the variables are separated in our equation, we perform integration to determine the function of temperature over time. The equation:$$\frac{dT}{T-20}=-0.125dt$$is integrated to yield:$$\int \frac{1}{T-20}dT=-0.125\int dt$$The result of this integration is the natural logarithm function:$$\ln |T-20|=-0.125t+C$$where $C$ is the constant of integration. Using the initial condition $T(0)=70$, we substitute back to find $C$. By solving $C$ and substituting back, we get the temperature as a function of time:$$T=50e^{-0.125t}+20$$This function demonstrates how the temperature decreases exponentially over time. The coffee's temperature continues to decline and approaches the ambient temperature ${20}^{\circ }C$ but never quite reaches it, illustrating the nature of exponential decay in cooling processes.