Question

Solve the differential equation for Newton's Law of Cooling to find the temperature in the following cases. Then answer any additional questions. A glass of milk is moved from a refrigerator with a temperature of \(5^{\circ} \mathrm{C}\) to a room with a temperature of \(20^{\circ} \mathrm{C}\). One minute later the milk has warmed to a temperature of \(7^{\circ} \mathrm{C}\). After how many minutes does the milk have a temperature that is \(90 \%\) of the ambient temperature?

Short answer

Answer: It takes approximately 2.84 minutes for the milk to reach 90% of the ambient temperature.

Step-by-step solution

Write down Newton's Law of Cooling equation

Newton's Law of Cooling states that the rate of change of temperature with respect to time, dT/dt, is proportional to the difference between the ambient temperature and the current temperature, that is, $$\frac{dT}{dt} = k(T_A - T)$$ where T is the temperature of the object, \(T_A\) is the ambient temperature, t is time, and k is a constant.

Solve the differential equation

Rewrite the cooling equation as a separable equation: $$\frac{1}{T_A - T} \frac{dT}{dt} = k$$ Solve the separable equation by integrating both sides with respect to time, $$\int \frac{1}{T_A - T} \frac{dT}{dt} dt = k dt$$ Integrate both sides: $$-\ln|T_A - T| = kt + C$$ Solve for T: $$T_A - T = e^{-kt - C}$$ which is equal to $$T = T_A - e^{-kt - C}$$

Find the particular solution

We can find the particular solution using the given initial conditions. The initial temperature of the milk is \(T(0) = 5^{\circ} C\) (when t = 0) and after 1 minute, the temperature is \(T(1) = 7^{\circ}C\). Ambient temperature is given as \(T_A = 20^{\circ} C\). Plug in \(T(0)\) and solve for C: $$5 = 20 - e^{-k(0) -C}$$ $$C = -\ln 15$$ Now, plug in \(T(1)\) and solve for k: $$7 = 20 - e^{-k(1) - (-\ln 15) }$$ $$\frac{13}{15} = e^{-k}$$ Take the natural logarithm of both sides: $$-\ln \frac{13}{15} = -k$$ So, $$k = \ln \frac{13}{15}$$

Calculate the time when milk reaches 90% of the ambient temperature

Now, we need to find the time 't' when the milk reaches 90% of the ambient temperature. Therefore, we have: $$0.9 \times 20 = 20 - e^{-kt}$$ or $$2 = e^{-kt}$$ Take the natural logarithm of both sides: $$\ln 2 = -k t$$ Now, insert the value of k we found earlier: $$\ln 2 = -(\ln \frac{13}{15})t$$ Finally, solve for t: $$t = \frac{\ln 2}{\ln \frac{15}{13}}$$ After calculating, we get approximately: $$t \approx 2.84 \, \text{minutes}$$. So, it takes around 2.84 minutes for the milk to reach 90% of the ambient temperature.

Key concepts

Differential Equations

A differential equation is like a blueprint that allows us to express a relationship between a function and its derivatives. This relationship is vital in describing phenomena where change is involved, such as the change in temperature over time. In the case of Newton's Law of Cooling, the differential equation shows us how the temperature of an object changes in relation to the ambient temperature around it.

For example, if you take a glass of milk out of the fridge, the rate at which its temperature changes depends on both the difference between its temperature and the room's temperature. The equation for this would be written as \(\frac{dT}{dt} = k(T_A - T)\), where \(T\) is the temperature of the milk, \(T_A\) the ambient temperature, and \(k\) a constant that characterizes the cooling process. Differential equations can often be complex, but they provide powerful tools for predicting how systems behave over time.

Separable Equations

In mathematics, a separable equation is a type of differential equation that can be manipulated so that all terms involving the dependent variable are on one side of the equation, and all terms involving the independent variable are on the other side.

For Newton's Law of Cooling, the equation \(\frac{dT}{dt} = k(T_A - T)\) can be rewritten in a separable form as \(\frac{1}{T_A - T} \frac{dT}{dt} = k\).
This makes it easier to solve because it allows us to integrate each side independently. By isolating separate variables, we can more easily apply calculus techniques to find the general solution, adding clarity to how variables interact over time.
Separable equations are particularly handy when we're dealing with natural processes, like cooling or growth, because they allow us to directly link cause (the influencing factors) with effect (the system's response).

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a special kind of logarithm based on the constant \(e\), which is approximately 2.71828. It is extensively used in calculus and in solving differential equations because it helps simplify exponential terms, making it easier to solve equations.

For our cooling problem, after converting the separable equation, we end up with an equation that involves \(-\ln|T_A - T| = kt + C\). This comes from integrating the left-hand side of the separable equation with respect to temperature, where \(C\) is the integration constant.
Taking the logarithm allows us to deal with the exponentiation involved effectively and turns the multiplicative problem into an additive one, which is far easier to manage and interpret in the context of real-world problems and predictions like cooling rates.

Integration

Integration is the process of finding the integral of a function, which is the reverse operation of differentiation. In the context of solving differential equations, integration helps us find the actual function that describes the system's behavior over time.

Once the differential equation is made separable and rearranged, we integrate both sides:

• The left side is integrated with respect to \(T\), resulting in \(-\ln|T_A - T|\).
• The right side is integrated with respect to \(t\), giving \(kt + C\), where \(C\) is the constant of integration.

This integration step is crucial because it allows us to express the temperature as an explicit function of time, \(T = T_A - e^{-kt - C}\). With this equation, we can now plug in any specific time value to predict the temperature, which is exactly what we do when trying to answer the problem about when milk's temperature reaches 90% of the ambient temperature. Integration connects the changes described by the differential equation into a comprehensive formula that models our situation.