Problem 33 - College Algebra

Question

Use the four-step procedure for solving variation problems given on page 424 to solve. The heat loss of a glass window varies jointly as the window's area and the difference between the outside and inside temperatures. A window 3 feet wide by 6 feet long loses 1200 Btu per hour when the temperature outside is \(20^{\circ}\) colder than the temperature inside. Find the heat loss through a glass window that is 6 feet wide by 9 feet long when the temperature outside is \(10^{\circ}\) colder than the temperature inside.

Step-by-Step Solution

Answer

The heat loss through a glass window that is 6 feet wide by 9 feet long when the temperature outside is \(10^{\circ}\) colder than the temperature inside is 1800 Btu per hour.

Step 1: Identify the Formula

In order to solve the problem, we need to establish the joint variation formula for heat loss. We know it is jointly proportional to the area and the temperature difference, so we will use the formula \(H = kAT\), where \(H\) is the heat loss, \(k\) is the constant of variation, \(A\) the area of the window, and \(T\) the temperature difference.

Step 2: Calculate the Constant of Variation

Substitute the given numbers into our formula to solve for \(k\). The heat loss \(H\) is 1200 Btu, the area \(A\) is \(3 ft \times 6 ft = 18 sq ft\), and the temperature difference \(T\) is \(20^{\circ}\). So, \(1200 = k \times 18 \times 20\). Solving for \(k\), we get \(k = \frac{1200}{(18 \times 20)} = 3.33 Btu/ (sq ft degree)\).

Step 3: Apply the Constant to the New Situation

Now we need to find out the heat loss of a window of \(6 ft \times 9 ft = 54 sq ft\) with the temperature difference being \(10^{\circ}\). Plugging these values into our formula, we have \(H = 3.33 \times 54 \times 10\), giving us a heat loss of \(H = 1800 Btu per hour\).

Key Concepts

Heat Loss

Heat loss through a window plays an essential role in understanding energy efficiency in buildings. It represents the amount of heat energy that escapes through the window. Generally measured in British thermal units (Btu) per hour, it can significantly affect indoor temperature regulation and energy costs.
Heat loss is influenced by several factors, including:

• The material of the window
• The thickness and type of glass
• The temperature difference between inside and outside
• The window's area

A clear understanding of heat loss helps in designing better insulation solutions for homes and offices, promoting energy conservation.

Area of Window

The area of a window directly affects how much heat it can potentially lose. In our context, it's the surface area through which heat escapes. The larger the area, the more significant the potential heat loss.
To find the area of the window, use the formula for the area of a rectangle:

• For a window 3 feet wide by 6 feet long, the area is:\(3 \times 6 = 18 \text{ square feet}\)
• For a 6 feet wide by 9 feet long window, the area is:\(6 \times 9 = 54 \text{ square feet}\)

The larger window in the new scenario suggests potentially more heat loss, assuming all other factors remain constant.

Temperature Difference

Temperature difference is a crucial factor in determining heat loss. It refers to the difference in temperature between the inside and outside of the window.
A greater temperature difference means more heat will flow from the warmer side to the cooler side, leading to increased heat loss.

• In the initial scenario, the temperature difference is \(20^{\circ} \text{F}\).
• In the new scenario, it is \(10^{\circ} \text{F}\).

This change in temperature difference affects the rate of heat loss. As the temperature difference decreases, the rate of heat loss typically decreases as well.

Variation Constant

The variation constant \(k\) represents the proportional relationship between heat loss, the area of the window, and the temperature difference. It's a unique value that helps to model how these factors interact.
In the given problem, you calculate this constant using known values of heat loss, area, and temperature difference:
\[k = \frac{\text{Heat Loss}}{\text{Area} \times \text{Temperature Difference}}\]
Using the first scenario's values:

• Heat loss \(H = 1200 \text{ Btu}\)
• Area \(A = 18 \text{ square feet}\)
• Temperature difference \(T = 20^{\circ}\)

The variation constant \(k = 3.33 \text{ Btu per square foot per degree}\). This constant allows you to predict heat loss in similar scenarios by applying it to new values of area and temperature difference.