Problem 12 - Chemistry The Molecular Science

Question

On a sunny day, solar energy reaches Earth at a rate of 4.0 J min \(^{-1} \mathrm{~cm}^{-2}\). Suppose a house has a square, flat roof of dimensions \(12 \mathrm{~m}\) by \(12 \mathrm{~m}\). Calculate how much solar energy reaches this roof in \(1.0 \mathrm{~h}\). (Note: This is why roofs painted with light- reflecting paint keep buildings cooler than black, unpainted roofs. The painted roofs reflect most of this energy rather than absorb it.)

Step-by-Step Solution

Answer

The solar energy reaching the roof in one hour is 345,600,000 joules.

Step 1: Convert Roof Dimensions to Square Centimeters

The roof is the area where solar energy reaches. The roof dimensions are given in meters but the rate of energy is given per square centimeter. First, convert the dimensions from meters to centimeters: \[ 12 \text{ m} = 1200 \text{ cm} \]The area of a square is calculated by squaring the side length:\[ \text{Area of roof} = 1200 \text{ cm} \times 1200 \text{ cm} = 1,440,000 \text{ cm}^2 \]

Step 2: Convert Time from Hours to Minutes

The solar energy rate is provided per minute. Convert the time from hours to minutes to match the units:\[ 1 \text{ hour} = 60 \text{ minutes} \]

Step 3: Calculate Total Solar Energy Received

Using the rate of solar energy and the total area, calculate the total energy received in one minute and then for one hour. First, find the energy per minute:\[ \text{Energy per minute} = 4.0 \text{ J/min/cm}^2 \times 1,440,000 \text{ cm}^2 = 5,760,000 \text{ J/min} \]Next, calculate for one hour (60 minutes):\[ \text{Total energy for 1 hour} = 5,760,000 \text{ J/min} \times 60 \text{ min} = 345,600,000 \text{ J} \]

Step 4: Check Units and Ensure Consistency

Verify that all units are consistent throughout the calculations. We have used centimeters for area and minutes for time consistently, thus ensuring the answer is in joules.

Key Concepts

Unit Conversion in Physics

Unit conversion is a fundamental concept in physics, especially when dealing with real-world problems where measurements can be in different units. In this exercise, we saw the need to convert roof dimensions from meters to centimeters and time from hours to minutes. This ensures consistency across our calculations.

When converting units, remember that each meter equals 100 centimeters. Thus, a measurement of 12 meters becomes 1200 centimeters. For time conversions, 1 hour is equivalent to 60 minutes. Adjusting these units helps simplify calculations and ensure they align with the given energy rate. By consistently using the correct units, errors and inaccuracies are minimized.

Key steps in unit conversion involve identifying the units given, determining the required unit, finding the conversion factor, and applying it mathematically. This process is essential for solving physics problems accurately.

Area Calculation

Knowing how to calculate the area is crucial in physics, particularly when determining how much of something affects a surface. In the given exercise, the roof's area was calculated to understand how much solar energy it could intercept.

The roof was specified as a square with a side length of 12 meters. Since 1 meter is equal to 100 centimeters, the side length in centimeters is 1200 cm. The area of a square is found by squaring its side length, resulting in an area of 1,440,000 square centimeters (cm²) for the roof.

Calculating area precisely is vital as it directly affects the end result in problems like energy absorption, where the amount of intercepted energy is a function of area. Remember that area always involves units of length squared, such as cm² or m².

Energy Absorption and Reflection

The concepts of energy absorption and reflection explain how surfaces interact with incoming energy. In the context of solar energy, surfaces like roofs play a critical role in determining how much energy a building absorbs versus how much is reflected away.

In the original problem, roofs with light-reflecting paint are mentioned as effective in reflecting most of the incident solar energy, keeping buildings cooler than black, unpainted roofs. This is because lighter surfaces reflect more sunlight, whereas darker surfaces absorb more energy, heating up the building.

The efficiency of energy absorption and reflection depends on surface type, texture, and color. Understanding these concepts helps in designing energy-efficient buildings that can maintain cooler interiors and reduce the need for air conditioning, saving energy.