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Chapter 4: Problem 82

Room Temperature Marlene brought a can of polyurethane varnish that was stored at \(40^{\circ} \mathrm{F}\) into her shop, where the temperature was \(74^{\circ} .\) After 2 hr the temperature of the varnish was \(58^{\circ} .\) If the varnish must be \(68^{\circ}\) for best results, then how much longer must Marlene wait until she uses the varnish?

Short Answer

Expert verified
Marlene should wait approximately 2.45 more hours.

Step by step solution

01

- Understand Newton鈥檚 Law of Cooling

Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature. The formula is given by \[T(t) = T_{\text{ambient}} + (T_{\text{initial}} - T_{\text{ambient}}) e^{-kt}\]where: - \(T(t)\) is the temperature at time \(t\), - \(T_{\text{ambient}}\) is the ambient temperature, - \(T_{\text{initial}}\) is the initial temperature, - \(k\) is the cooling constant, - \(t\) is the time.
02

- Identify the known values

Given in the problem: - Initial temperature \(T_{\text{initial}} = 40^{\bullet}F\) - Ambient temperature \(T_{\text{ambient}} = 74^{\bullet}F\) - Temperature after \(2\) hours \(T(2) = 58^{\bullet} F\)
03

- Set up the equation for temperature after 2 hours

The temperature after \(2\) hours is given by: \[58 = 74 + (40 - 74) e^{-2k}\]Simplify the equation to find \(k\).
04

- Solve for the cooling constant k

\[58 - 74 = (40 - 74) e^{-2k}\]\[-16 = -34 e^{-2k}\]\[e^{-2k} = \frac{16}{34}\]\[e^{-2k} = \frac{8}{17}\]Take the natural logarithm on both sides to solve for \(k\): \[-2k = \text{ln} \frac{8}{17}\]\[k = -\frac{1}{2} \text{ln} \frac{8}{17}\]
05

- Set up the equation for the desired temperature

We need to find the time when the temperature is \(68^{\bullet} F\):\[68 = 74 + (40 - 74) e^{-kt}\]\[68 - 74 = (40 - 74) e^{-kt}\]\[-6 = -34 e^{-kt}\]\[e^{-kt} = \frac{6}{34}\]\[e^{-kt} = \frac{3}{17}\]
06

- Solve for the time t

Using the natural logarithm:\[-kt = \text{ln} \frac{3}{17}\]Substitute \(k\) from Step 4 into the equation:\[\left(-\frac{1}{2} \text{ln} \frac{8}{17} \right)t = \text{ln} \frac{3}{17}\]Solve for \(t\):\[t = \frac{\text{ln} \frac{3}{17}}{-\frac{1}{2} \text{ln} \frac{8}{17}}\]
07

- Calculate the time t

Using a calculator to find the natural logarithms:\[\text{ln} \frac{3}{17} \text{ and } \text{ln} \frac{8}{17}\]Let \[\text{ln} \frac{3}{17} \text{ approximately = -1.7346}\] \[n\text{ln} \frac{8}{17} \text{ approximately = -0.7793}\]Thus, \[t \approx \frac{-1.7346}{-0.7793/ 2} \approx 4.45 \text{ hrs}\]
08

- Calculate the additional time Marlene should wait

Since 2 hours have already passed, the additional time needed is: \[4.45 - 2 = 2.45 \text{ hours}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

temperature change
Understanding temperature change is key to mastering Newton's Law of Cooling. When an object is placed in a different temperature environment, the change in its temperature happens gradually. This process is influenced by the difference between the object鈥檚 temperature and the surrounding (ambient) temperature. The larger the difference, the faster the rate of temperature change initially. Over time, as the temperature gap narrows, the cooling or heating rate slows down.
cooling constant
The cooling constant, denoted as k in the formula, is a measure of how quickly the object approaches the ambient temperature. It is unique to the material and conditions of the object. A large k value means the temperature changes quickly, while a small k value indicates a slower change. The cooling constant can be determined using observational data, like the temperature change after a certain period. In our solution, we calculated k based on the temperature change from 40掳F to 58掳F over 2 hours. The formula involving the natural logarithm \(ln\) helps in isolating and solving for k.
ambient temperature
Ambient temperature refers to the temperature of the surrounding environment where the object is placed. This temperature acts as a base level that the object鈥檚 temperature will asymptotically approach over time. In the given problem, Marlene's workshop has an ambient temperature of 74掳F. It's crucial to know this value because it sets the target for the cooling or warming process. The object will never surpass the ambient temperature naturally in cooling or heating without external influence.
exponential decay
Exponential decay describes how quantities reduce over time and is a central concept in Newton鈥檚 Law of Cooling. The formula \(e^{-kt}\) expresses this decay, showing how the temperature difference between the object and the ambient environment decreases exponentially. This means the rate of temperature change decreases over time, making significant changes early on, but slower changes later. Understanding exponential decay helps in predicting how long it will take for an object to reach a desired temperature. For instance, calculating when the varnish temperature reaches 68掳F involved solving an exponential decay equation using logarithms.

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Most popular questions from this chapter

Consider the function \(y=\log \left(10^{n} \cdot x\right)\) where \(n\) is an integer. Use a graphing calculator to graph this function for several choices of \(n .\) Make a conjecture about the relationship between the graph of \(y=\log \left(10^{n} \cdot x\right)\) and the graph of \(y=\log (x) .\) Save your conjecture and attempt to prove it after you have studied the properties of logarithms, which are coming in Section \(4.3 .\) Repeat this exercise with \(y=\log \left(x^{n}\right)\) where \(n\) is an integer.

Solve each equation. Find the exact solutions. $$e^{3 x-4}=1$$

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Find the equations of the horizontal and vertical asymptotes for the graph of the function \(f(x)=\frac{2 x-3}{x+7}\)

To evaluate an exponential or logarithmic function we simply press a button on a calculator. But what does the calculator do to find the answer? The next exercises show formulas from calculus that are used to evaluate \(e^{x}\) and \(\ln (1+x)\). Infinite Series for \(e^{x}\) The following formula from calculus is used to compute values of \(e^{x}\) : $$e^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\cdots+\frac{x^{n}}{n !}+\cdots$$ where \(n !=1 \cdot 2 \cdot 3 \cdot \cdots \cdot n\) for any positive integer \(n .\) The notation \(n !\) is read " \(n\) factorial." For example, \(3 !=1 \cdot 2 \cdot 3=6\) In calculating \(e^{x},\) the more terms that we use from the formula, the closer we get to the true value of \(e^{x}\). Use the first five terms of the formula to estimate the value of \(e^{0.1}\) and compare your result to the value of \(e^{0.1}\) obtained using the \(e^{x}-\) key on your calculator.
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