Problem 28
Question
Your home loses energy in the winter through doors, windows, and any poorly insulated walls. A sliding glass door \((6 \mathrm{ft} \times 6.5 \mathrm{ft} \text { with } 0.5\) inch of insulating glass ) allows \(1.0 \times 10^{6} \mathrm{J} / \mathrm{h}\) to pass through the glass if the inside temperature is \(22^{\circ} \mathrm{C}\left(72^{\circ} \mathrm{F}\right)\) and the outside temperature is \(0^{\circ} \mathrm{C}\left(32^{\circ} \mathrm{F}\right) .\) What quantity of energy, expressed in kilojoules, is lost per day? Assume that your house is heated by electricity. How many kilowatt-hours of energy are lost per day through the door? (See Study Question 24 for the definition of a kilowatt- hour.
Step-by-Step Solution
Verified Answer
24000 kJ/day and approximately 6.67 kWh are lost per day.
1Step 1: Calculate Energy Lost Per Hour in Kilojoules
The energy loss given is in joules per hour. First, we need to convert this to kilojoules. Since 1 kilojoule (kJ) is 1000 joules (J), we can express the hourly energy loss through the door in kilojoules:\[1.0 \times 10^{6} \text{ J/h} = \frac{1.0 \times 10^{6}}{1000} \text{ kJ/h} = 1000 \text{ kJ/h}\]
2Step 2: Calculate Energy Lost Per Day in Kilojoules
Since the energy loss per hour through the door is 1000 kJ, we multiply this by 24 hours to find the total energy lost per day:\[1000 \text{ kJ/h} \times 24 \text{ h} = 24000 \text{ kJ/day}\]
3Step 3: Calculate Energy Lost Per Day in Kilowatt-hours
One kilowatt-hour (kWh) is equivalent to 3600 kJ (since 1 kWh = 1000 W \(\times\) 3600 s). To find out how many kilowatt-hours of energy are lost per day, we divide the total daily energy loss in kilojoules by 3600:\[\frac{24000 \text{ kJ}}{3600 \text{ kJ/kWh}} \approx 6.67 \text{ kWh}\]
Key Concepts
InsulationKilojoule ConversionKilowatt-hourTemperature Differential
Insulation
Insulation plays a critical role in minimizing energy loss in a home, particularly in cold weather. It acts as a barrier to heat flow and helps in maintaining a comfortable indoor temperature. The level of insulation impacts how much energy is required to heat or cool a house. The better the insulation, the less energy is lost to the outside environment.
When dealing with insulated parts of a house, like the sliding glass door mentioned in the exercise, the level of insulation can significantly affect energy efficiency. Here, the door has 0.5 inch of insulating glass, which helps reduce energy loss but is not entirely effective against the temperature difference between the inside and outside.
Repeated heat loss through weakly insulated areas means more energy, and thus more cost, to maintain desired indoor temperatures. This is why understanding insulation properties and considering improvements can be both economical and environmentally beneficial.
When dealing with insulated parts of a house, like the sliding glass door mentioned in the exercise, the level of insulation can significantly affect energy efficiency. Here, the door has 0.5 inch of insulating glass, which helps reduce energy loss but is not entirely effective against the temperature difference between the inside and outside.
Repeated heat loss through weakly insulated areas means more energy, and thus more cost, to maintain desired indoor temperatures. This is why understanding insulation properties and considering improvements can be both economical and environmentally beneficial.
Kilojoule Conversion
Converting energy from one unit to another is vital in understanding and calculating energy consumption. One common conversion is from joules, the fundamental unit of energy in the International System of Units, to kilojoules.
The conversion is straightforward:
The conversion is straightforward:
- 1 kilojoule (kJ) = 1000 joules (J).
Kilowatt-hour
The kilowatt-hour (kWh) is a unit of energy commonly used to measure electricity consumption. Understanding this unit is essential when discussing electrical energy usage because it's used for billing by power companies.
1 kilowatt-hour is defined as the consumption of 1000 watts of energy over one hour, which is equivalent to 3600 kilojoules.
In the given exercise, calculating how many kWh of energy are lost per day through the glass door involves dividing the daily energy loss in kilojoules by 3600, resulting in approximately 6.67 kWh. This conversion provides insight into the actual cost implications of energy loss through the door, connecting physical concepts to real-world utility bills.
1 kilowatt-hour is defined as the consumption of 1000 watts of energy over one hour, which is equivalent to 3600 kilojoules.
In the given exercise, calculating how many kWh of energy are lost per day through the glass door involves dividing the daily energy loss in kilojoules by 3600, resulting in approximately 6.67 kWh. This conversion provides insight into the actual cost implications of energy loss through the door, connecting physical concepts to real-world utility bills.
Temperature Differential
The temperature differential between two environments significantly influences the rate of energy transfer between them. The greater the difference, the faster energy (heat) will flow from the warmer to the colder environment.
In the exercise example, the inside temperature is 22°C (72°F), while the outside is 0°C (32°F). This creates a temperature differential of 22°C, which drives heat through the sliding glass door despite the insulation, resulting in the calculated energy loss.
Understanding this concept is essential for assessing the effectiveness of insulation materials and strategies. The larger the temperature difference, the more robust the insulation needed to minimize energy loss effectively.
In the exercise example, the inside temperature is 22°C (72°F), while the outside is 0°C (32°F). This creates a temperature differential of 22°C, which drives heat through the sliding glass door despite the insulation, resulting in the calculated energy loss.
Understanding this concept is essential for assessing the effectiveness of insulation materials and strategies. The larger the temperature difference, the more robust the insulation needed to minimize energy loss effectively.
Other exercises in this chapter
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