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A degree-day is a unit used to measure the fuel requirements of buildings. By definition, each degree that the average daily temperature is below \(65^{\circ} \mathrm{F}\) is 1 degree-day. For example, an average daily temperature of \(42^{\circ} \mathrm{F}\) constitutes 23 degree-days. If the average temperature on January 1 was \(42^{\circ} \mathrm{F}\) and fell \(2^{\circ} \mathrm{F}\) for each subsequent day up to and including January 10 , how many degree-days are included from January 1 to January \(10 ?\)

Short Answer

Expert verified
The total number of degree-days from January 1 to January 10 is 330.

Step by step solution

01

Identify the temperature for each day

The temperature on January 1 is given as \(42^{\circ} \mathrm{F}\). It falls \(2^{\circ} \mathrm{F}\) each day, so the temperatures for the 10 days will form an arithmetic sequence: \(42, 40, 38, ..., 22^{\circ} \mathrm{F}\).
02

Calculate degree-days for each day

A degree-day is each degree that the average daily temperature is below \(65^{\circ} \mathrm{F}\). Therefore, for each day, subtract the day's temperature from \(65^{\circ} \mathrm{F}\). This will also form an arithmetic sequence: \(23, 25, 27, ..., 43\) degree-days.
03

Sum up the degree-days

Finally, to find the total degree-days from January 1 to January 10, sum up the degree-days from each day. You can use the formula for the sum of an arithmetic progression: \(S = n/2 * (a + l)\) where \(n = 10\) is the number of terms (days), \(a = 23\) and \(l = 43\) are the first and last term of the sequence. So, \(S = 10/2 * (23 + 43) = 330\).

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Key Concepts

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

Arithmetic Sequence
An arithmetic sequence is a list of numbers where each term after the first is found by adding a constant, called the common difference, to the previous term. In our context, the temperature on January 1 starts at \(42^{\text{o}} \text{F}\) and decreases by \(2^{\text{o}} \text{F}\) each day, creating an arithmetic sequence of temperatures: \(42, 40, 38, ..., 22^{\text{o}} \text{F}\). Understanding how to work with arithmetic sequences is pivotal when calculating degree-days over a period because it allows you to predict temperatures for each day without measuring them directly.

For example, if you know the common difference (in this case, the daily temperature drop) and the initial temperature, the temperature for any subsequent day can be easily determined. This understanding is accelerated by using the arithmetic sequence formula for the nth term: \(a_n = a_1 + (n-1)d\), where \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the term number in the sequence.
Average Daily Temperature
The concept of average daily temperature is crucial for understanding degree-days. It represents the mean temperature of a day, which is essential for calculating the heating or cooling needs of a building. In our case, the average daily temperature starts at \(42^{\text{o}} \text{F}\) and falls each day. Because buildings typically require heating when the temperature drops below \(65^{\text{o}} \text{F}\), we use this benchmark to determine the number of degree-days.

By comparing the average daily temperature to a base temperature of \(65^{\text{o}} \text{F}\), we come up with a simple measure of heating demand. Each degree of temperature below the base counts as one degree-day. In warmer climates or seasons, cooling degree-days can also be calculated by considering temperatures above a certain base.
Fuel Requirements Measurement
Degree-days serve as a unit of measurement for the fuel requirements of buildings, essentially quantifying the demand for heating. They accumulate over time, creating a practical metric for energy consumption forecasting and cost estimation for heating buildings. The larger the number of degree-days, the colder the weather has been, and likely, more fuel was required to maintain comfortable indoor temperatures.

By summing the degree-days over a period, as demonstrated with the arithmetic sequence in our exercise, we can estimate fuel usage over that time. The degree-day method is especially significant for energy providers and planners to anticipate fuel needs and for building managers in budgeting for heating costs. This reliable method allows for comparing different periods or locations with variable temperature patterns, making it an indispensable tool in energy management and conservation efforts.

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Most popular questions from this chapter

Follow the outline below and use mathematical induction to prove the Binomial Theorem: $$\begin{aligned}(a+b)^{n} &-\left(\begin{array}{c}n \\\0\end{array}\right) a^{n}+\left(\begin{array}{c}n \\\1\end{array}\right) a^{n-1} b+\left(\begin{array}{c}n \\\2\end{array}\right) a^{n-2} b^{2} \\\&+\cdots+\left(\begin{array}{c}n \\\n-1\end{array}\right) a b^{n-1}+\left(\begin{array}{c}n \\\n\end{array}\right) b^{n}\end{aligned}$$ a. Verify the formula for \(n-1\) b. Replace \(n\) with \(k\) and write the statement that is assumed true. Replace \(n\) with \(k+1\) and write the statement that must be proved. c. Multiply both sides of the statement assumed to be true by \(a+b .\) Add exponents on the left. On the right, distribute \(a\) and \(b,\) respectively. d. Collect like terms on the right. At this point, you should have $$\begin{array}{l}(a+b)^{k+1}-\left(\begin{array}{c}k \\\0\end{array}\right)a^{k+1}+\left[\left(\begin{array}{c}k \\\0\end{array}\right)+\left(\begin{array}{c}k \\\1\end{array}\right)\right] a^{k} b \\\\+\left[\left(\begin{array}{c}k \\\1\end{array}\right)+\left(\begin{array}{c}k \\\2\end{array}\right)\right] a^{k-1} b^{2}+\left[\left(\begin{array}{c}k \\\2\end{array}\right)+\left(\begin{array}{c}k \\\3\end{array}\right)\right] a^{k-2} b^{3} \\\\+\cdots+\left[\left(\begin{array}{c}k \\\k-1\end{array}\right)+\left(\begin{array}{c}k \\\k\end{array}\right)\right] a b^{k}+\left(\begin{array}{c}k \\\k\end{array}\right) b^{k+1}\end{array}$$ e. Use the result of Exercise 84 to add the binomial sums in brackets. For example, because \(\left(\begin{array}{l}n \\\ r\end{array}\right)+\left(\begin{array}{c}n \\ r+1\end{array}\right)\) $$\begin{aligned}&-\left(\begin{array}{l}n+1 \\\r+1\end{array}\right), \text { then }\left(\begin{array}{l}k \\\0\end{array}\right)+\left(\begin{array}{l}k \\\1\end{array}\right)-\left(\begin{array}{c}k+1 \\\1\end{array}\right) \text { and }\\\&\left(\begin{array}{l}k \\\1\end{array}\right)+\left(\begin{array}{l}k \\\2\end{array}\right)-\left(\begin{array}{c}k+1 \\\2\end{array}\right)\end{aligned}$$ f. Because \(\left(\begin{array}{l}k \\\ 0\end{array}\right)-\left(\begin{array}{c}k+1 \\ 0\end{array}\right)(\text { why? })\) and \(\left(\begin{array}{l}k \\\ k\end{array}\right)-\left(\begin{array}{l}k+1 \\ k+1\end{array}\right)\) (why?), substitute these results and the results from part (e) into the equation in part (d). This should give the statement that we were required to prove in the second step of the mathematical induction process.

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