Question

A body in a room at \(60^{\circ}\) cools from \(200^{\circ}\) to \(120^{\circ}\) in halfan hour. (a) Show that its tmperature after \(\mathrm{t}\) minutes is \(60+140 \mathrm{e}^{-\mathrm{lt}}\), where \(\mathrm{k}=(\ln 7-\ln 3) / 30\) (b) Show that the time \(t\) required to reach a temperature of \(\mathrm{T}\) degrees is given by the formula \(\mathrm{t}=[\ln 140-\ln (\mathrm{T}-60)] / \mathrm{k}\), where \(60<\mathrm{T} \leq 200\). (c) Find the time at which the temperature is \(90^{\circ}\). (d) Find a formula for the temperature of the body at time \(t\) if the room temperature is not kept constant but falls at a rate of \(1^{\circ}\) each ten minutes. Assume the room temperature is \(60^{\circ}\) when the body temperature is \(200^{\circ}\).

Short answer

Question: Using Newton's Law of Cooling, determine the approximate time it takes for a body to cool down to 90°C if initially, it is at 200°C and cools down to 120°C in 30 minutes in a room with a constant temperature of 60°C. Answer: It takes approximately 15.29 minutes for the body to cool down to 90°C.

Step-by-step solution

Setup Newton's Law of Cooling

Newton's Law of Cooling states that the rate at which the temperature of a body changes is proportional to the difference between the body's temperature and the surrounding environment (room) temperature. Mathematically, this can be written as: $$ \frac{dT}{dt} = -k(T - T_r) $$ Where: - \(T\) is the temperature of the body, - \(T_r\) is the temperature of the room (constant), - \(k\) is the cooling constant.

Solve the Differential Equation

Since we have a first-order linear differential equation, we can solve it directly. Separating the variables and integrating, we get: $$ \int \frac{dT}{T-T_r} = -k \int dt $$ After integrating, we get: $$ \ln |T - T_r| = -kt + C $$ Where \(C\) is the constant of integration.

Solve for the Temperature Equation

Now, we will find the equation for the temperature of the body at any given time \(t\). We can get rid of the natural logarithm by taking the exponential of both sides: $$ T - T_r = e^{-kt + C} $$ Let's rearrange the equation for \(T\): $$ T = T_r + e^{C}e^{-kt} $$

Determine the Initial Temperature

We are given that the initial temperature of the body is \(200^{\circ}\). So, when \(t = 0\), \(T = 200^{\circ}\). Plugging these values into our temperature equation, we get: $$ 200 = 60 + e^{C}e^{0} $$ This simplifies to: $$ e^{C} = 140 $$

Calculate the Cooling Constant, \(k\)

Next, we need to find the cooling constant, \(k\). We are given that the body cools down to \(120^{\circ}\) in 30 minutes. Plugging these values into our temperature equation, we get: $$ 120 = 60 + 140e^{-30k} $$ This simplifies to: $$ e^{-30k} = \frac{60}{140} = \frac{3}{7} $$ Taking the natural logarithm of both sides, we arrive at: $$ -30k = \ln \frac{3}{7} $$ Finally, solving for \(k\), we find: $$ k = \frac{\ln 7 - \ln 3}{30} $$

Find the Final Temperature Formula

Now that we have found \(e^C\) and \(k\), we can plug their values into the temperature equation to get the final formula: $$ T(t) = 60 + 140e^{-(\frac{\ln 7 - \ln 3}{30})t} $$ This is the temperature of the body after \(t\) minutes.

Time to Reach Temperature \(T\)

Now, let's find the time required to reach a temperature \(T\). We will solve for \(t\) in the temperature equation: $$ T = 60 + 140e^{-(\frac{\ln 7 - \ln 3}{30})t} $$ Rearranging, $$ e^{-(\frac{\ln 7 - \ln 3}{30})t} = \frac{T - 60}{140} $$ Taking the natural logarithm of both sides, $$ -(\frac{\ln 7 - \ln 3}{30})t = \ln (\frac{T - 60}{140}) $$ Finally, solving for \(t\), we get: $$ t = \frac{\ln 140 - \ln (T - 60)}{\frac{\ln 7 - \ln 3}{30}} $$

Find the Time to Reach \(90^{\circ}\)

We want to find the time it takes to reach a temperature of \(90^{\circ}\). Plugging \(T = 90\) into the time equation, we get: $$ t = \frac{\ln 140 - \ln (90 - 60)}{\frac{\ln 7 - \ln 3}{30}} $$ Calculating the value, we get: $$ t \approx 15.29 \ \text{minutes} $$

Find the Formula When Room Temperature Decreases

The room temperature decreases by \(1^{\circ}\) every 10 minutes. So, we can change our room temperature \(T_r\) to accommodate this new information: $$ T_r(t) = 60 - \frac{t}{10} $$ We will rewrite the differential equation with the new \(T_r(t)\): $$ \frac{dT}{dt} = -k(T - T_r(t)) $$ This differential equation is more difficult to solve analytically due to the presence of the time variable in the room temperature. We would need either to approximate a solution or to use numerical methods.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They are fundamental to understanding various phenomena in sciences and engineering, from the flight path of an airplane to the fluctuation of stock markets. In the context of Newton's Law of Cooling, a differential equation models how the temperature of an object changes over time in relation to the temperature of its surroundings.

Specifically, the differential equation \( \frac{dT}{dt} = -k(T - T_r) \) expresses that the rate of change of the temperature \(T\) with respect to time \(t\) is proportional to the difference between the object's temperature and the room temperature \(T_r\). The minus sign indicates the temperature of the object decreases over time if it is higher than the room temperature. Once we set up this equation, we can solve it by integrating both sides, as shown in the textbook solution. This process transforms our equation from involving rates of change to one that directly relates temperature to time.

Exponential Decay

Exponential decay describes the process in which a quantity decreases at a rate proportional to its current value. In the case of cooling, the temperature's rate of decrease is proportional to the difference between the object's current temperature and the ambient temperature, indicating an exponential relationship.

After integrating the differential equation set by Newton's Law of Cooling, we obtain an expression featuring an exponential function, manifesting as \( e^{-kt} \). This shows that the temperature change over time follows an exponential decay pattern. For instance, as \(t\) grows, \(e^{-kt}\) becomes smaller, signifying that the object is cooling down. The solution demonstrates that when the object cools from \(200^\(circ\)\) to \(120^\(circ\)\) within half an hour, this decay can be quantified and the exact temperature at any given moment can be predicted.

Cooling Constant

The cooling constant \(k\) in Newton's Law of Cooling is a crucial factor that quantitatively describes how fast an object cools down in a particular environment. It is determined by the material properties of the object, such as its heat capacity and thermal conductivity, and by the characteristics of the surrounding medium.

To calculate \(k\), we use information from a specific point in time — when the object has cooled to a known temperature after a measured period. The solution guides us through the calculations, using given temperature readings at two time points to find \(k\). Once \(k\) is determined, it remains constant for the cooling process under consistent conditions, allowing us to model the temperature of the object at any future time accurately. The concept of the cooling constant is pivotal because it embodies all the substance-specific and environmental factors into a single representative value.