Problem 22
Question
A beam of unknown temperature An aluminum beam was brought from the outside cold into a machine shop where the temperature was held at \(65^{\circ} \mathrm{F}\) . After 10 \(\mathrm{min}\) , the beam warmed to \(35^{\circ} \mathrm{F}\) and after another 10 \(\mathrm{min}\) it was \(50^{\circ} \mathrm{F}\) . Use Newton's law of cooling to estimate the beam's initial temperature.
Step-by-Step Solution
Verified Answer
The initial temperature of the beam was 5°F.
1Step 1: Understand 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 its own temperature and the ambient temperature. Mathematically, the law is represented as: \[\frac{dT}{dt} = k(T - T_a),\] where \(T\) is the temperature of the object at time \(t\), \(T_a\) is the ambient temperature, and \(k\) is a constant.
2Step 2: Define the Problem Variables
Let \(T_0\) be the initial temperature of the beam. After 10 minutes, the temperature of the beam is 35°F, and after 20 minutes, it is 50°F. The ambient temperature, \(T_a\), is 65°F.
3Step 3: Set Up the Cooling Equation at 10 Minutes
Using the formula for temperature over time under Newton's Law of Cooling, we write:\[ T(t) = T_a + (T_0 - T_a)e^{kt}. \] At \(t = 10\) minutes, \(T(10) = 35\). Substitute the values:\[ 35 = 65 + (T_0 - 65)e^{10k}. \] This equation will help us determine the relationship between \(T_0\) and \(k\).
4Step 4: Set Up the Cooling Equation at 20 Minutes
Similarly, at \(t = 20\) minutes, \(T(20) = 50\). Substitute the values:\[ 50 = 65 + (T_0 - 65)e^{20k}. \] We will use this equation along with the one from Step 3 to solve for \(T_0\) and \(k\).
5Step 5: Solve for the Decay Constant (k)
From Step 3, rearranging gives:\[ e^{10k} = \frac{35 - 65}{T_0 - 65} = \frac{-30}{T_0 - 65}. \] From Step 4, rearranging gives:\[ e^{20k} = \frac{50 - 65}{T_0 - 65} = \frac{-15}{T_0 - 65}. \] Divide the second equation by the first to eliminate \(T_0\):\[ \left(\frac{-15}{-30}\right) = \left(\frac{-15}{T_0 - 65}\right) \div \left(\frac{-30}{T_0 - 65}\right) = e^{10k}. \] Thus, \[ \frac{1}{2} = e^{10k}. \] Taking the natural logarithm gives:\[ 10k = \ln\left(\frac{1}{2}\right), \] so \[ k = \frac{\ln\left(\frac{1}{2}\right)}{10}. \]
6Step 6: Solve for Initial Temperature (T_0)
Using the value of \(k\) found in the previous step, substitute back into the equation from Step 3:\[ e^{10k} = \frac{-30}{T_0 - 65}. \] Substitute \(e^{10k} = \frac{1}{2}\), resulting in:\[ \frac{1}{2} = \frac{-30}{T_0 - 65}. \] Solving for \(T_0\) gives:\[ T_0 - 65 = -60, \] so \[ T_0 = 5. \]
Key Concepts
Temperature DecayAmbient TemperatureExponential Decay ModelThermal Physics
Temperature Decay
When we refer to **temperature decay**, we are talking about how the temperature of an object decreases over time towards the ambient temperature of its surroundings. This is a gradual process and depends heavily on the environment around the object.
In the context of Newton's Law of Cooling, temperature decay can be modeled mathematically. Essentially, it describes how an object's initial temperature continuously lowers until it reaches the temperature of the ambient environment.
Key aspects to consider about temperature decay include:
In the context of Newton's Law of Cooling, temperature decay can be modeled mathematically. Essentially, it describes how an object's initial temperature continuously lowers until it reaches the temperature of the ambient environment.
Key aspects to consider about temperature decay include:
- The initial temperature - the starting temperature of the object.
- The ambient temperature - the temperature of the surrounding environment.
- The decay constant - a constant that dictates the rate at which the temperature decreases.
Ambient Temperature
**Ambient temperature** is simply the surrounding temperature in a particular location or environment where an object is situated. For any object undergoing a change in temperature, the ambient temperature is a key factor that influences how rapidly or slowly the object can cool or warm.
In thermal processes like those described by Newton's Law of Cooling, the ambient temperature acts as the asymptotic limit towards which the object's temperature will eventually settle.
Some key points about ambient temperature:
In thermal processes like those described by Newton's Law of Cooling, the ambient temperature acts as the asymptotic limit towards which the object's temperature will eventually settle.
Some key points about ambient temperature:
- It serves as a constant reference point that does not change, unlike the object's temperature.
- For cooling processes, the object's temperature will always move towards the ambient temperature over time.
- Ambient temperature is crucial in determining whether an object will ultimately cool down or warm up.
Exponential Decay Model
The **exponential decay model** is a mathematical representation used to describe processes where quantities decrease at a rate proportional to their current value. This model is widely applicable in physics, including in describing how temperatures change over time.
More specifically, in Newton's Law of Cooling, the exponential decay model is used to calculate how quickly or slowly an object will reach the ambient temperature. The model is generally expressed with the equation:\[ T(t) = T_a + (T_0 - T_a)e^{kt} \]where:
More specifically, in Newton's Law of Cooling, the exponential decay model is used to calculate how quickly or slowly an object will reach the ambient temperature. The model is generally expressed with the equation:\[ T(t) = T_a + (T_0 - T_a)e^{kt} \]where:
- T(t) is the temperature at time t,
- T_0 is the initial temperature,
- T_a is the ambient temperature,
- k is the decay constant.
Thermal Physics
**Thermal physics** is a branch of physics that deals with heat, temperature, and their interrelation with energy and work. This field of science helps us to understand how energy in the form of heat is transferred and transformed.
Within thermal physics, Newton's Law of Cooling is a key concept used to describe the cooling (or heating) of an object. This principle focuses on how temperature changes depend on the material properties and the surrounding environment.
Here are some fundamentals of thermal physics:
Here are some fundamentals of thermal physics:
- Heat Transfer: The movement of heat from one body or material to another. This can occur through conduction, convection, or radiation.
- Specific Heat Capacity: A property that defines how much energy is needed to change the temperature of a substance.
- Thermal Equilibrium: A state in which two objects interacting do not exchange heat energy because they are at the same temperature.
Other exercises in this chapter
Problem 22
In Exercises \(13-24,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\ln \sinh v-\frac{1}{2} \operatorname{coth}^{2} v $$
View solution Problem 22
In Exercises \(17-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=(1+2 x) e^{-2 x} $$
View solution Problem 22
In Exercises \(5-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\frac{x \ln x}{1+\ln x} $$
View solution Problem 22
Find the derivative of \(y\) with respect to the given independent variable. \(y=5^{-\cos 2 t}\)
View solution