/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 A wall in a house contains a sin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A wall in a house contains a single window. The window consists of a single pane of glass whose area is \(0.16 \mathrm{m}^{2}\) and whose thickness is \(2.0 \mathrm{mm}\). Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are \(18 \mathrm{m}^{2}\) and \(0.10 \mathrm{m},\) respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window?

Short Answer

Expert verified
The window loses approximately 92.2% of the total heat.

Step by step solution

01

Formula for Heat Conduction

The heat lost through a material by conduction can be calculated using the formula: \( Q = \frac{k \cdot A \cdot \Delta T \cdot t}{d} \) where \( Q \) is the heat transferred, \( k \) is the thermal conductivity of the material, \( A \) is the area, \( \Delta T \) is the temperature difference across the material, \( t \) is time, and \( d \) is the thickness of the material.
02

Thermal Conductivities of Materials

To find the heat lost through the window and wall, we need to know the thermal conductivity \( k \) of glass and Styrofoam. Typically, \( k \) for glass is about 0.8 W/m·K, and for Styrofoam, it is about 0.03 W/m·K.
03

Calculate Heat Loss Through the Window

Using the formula from Step 1 for the window:\[ Q_{window} = \frac{0.8 \cdot 0.16 \cdot \Delta T \cdot t}{0.002} \]Simplifying:\[ Q_{window} = 64 \cdot \Delta T \cdot t \]
04

Calculate Heat Loss Through the Wall

Using the formula from Step 1 for the wall:\[ Q_{wall} = \frac{0.03 \cdot 18 \cdot \Delta T \cdot t}{0.1} \]Simplifying:\[ Q_{wall} = 5.4 \cdot \Delta T \cdot t \]
05

Total Heat Loss Calculation

Add the heat loss through the window and the wall:\[ Q_{total} = Q_{window} + Q_{wall} = 64 \cdot \Delta T \cdot t + 5.4 \cdot \Delta T \cdot t \]Simplifying:\[ Q_{total} = 69.4 \cdot \Delta T \cdot t \]
06

Calculate Percentage Heat Loss Through the Window

The percentage of total heat loss through the window is:\[ \frac{Q_{window}}{Q_{total}} \times 100 = \frac{64 \cdot \Delta T \cdot t}{69.4 \cdot \Delta T \cdot t} \times 100 \]This simplifies to:\[ \frac{64}{69.4} \times 100 \approx 92.2\% \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Thermal Conductivity
The thermal conductivity of a material is a measure of its ability to conduct heat. It is represented by the symbol \( k \) and is usually expressed in units of watts per meter per Kelvin (\( ext{W/m·K} \)). This property tells us how quickly or slowly heat can pass through a material. Materials with high thermal conductivity, like metals, quickly transfer heat, while materials with low thermal conductivity, such as Styrofoam, are great insulators.

In our exercise, the thermal conductivity of glass is about 0.8 W/m·K, which means it conducts heat relatively easily compared to Styrofoam, which has a low thermal conductivity of about 0.03 W/m·K.

  • High \( k \): Good at conducting heat, e.g., metals.
  • Low \( k \): Good insulators, e.g., Styrofoam.
Understanding thermal conductivity is crucial for designing buildings, as it helps determine how much insulation is needed to minimize heat loss.
Heat Transfer
Heat transfer is the movement of heat from a hotter area to a cooler one. In the context of our exercise, heat is lost from the inside of the house to the outside, primarily through conduction. The rate of heat transfer is dependent on several factors, including the thermal conductivity of the materials involved, their surface area, and thickness.

Using the formula for conduction, \( Q = \frac{k \cdot A \cdot \Delta T \cdot t}{d} \), we can calculate how much heat is being transferred. Here, \( A \) is the area through which heat is being transferred, \( \Delta T \) is the temperature difference, and \( d \) is the thickness of the material through which heat is conducted.

  • Area \( A \) affects how much space is available for heat to pass through.
  • Thickness \( d \) affects how long it takes for heat to pass through.
  • Temperature difference \( \Delta T \) affects the direction and intensity of heat flow.
Efficient heat transfer is essential for designing energy-efficient homes, reducing energy usage, and maintaining comfortable indoor environments.
Insulation Materials
Insulation materials are essential in reducing unwanted heat loss or gain by creating a barrier between areas of different temperatures. Such materials are key in household energy efficiency, as they help retain warmth in winter and cool air in summer.

In our example, Styrofoam is used as an insulating material for the wall. With a very low thermal conductivity of 0.03 W/m·K, it efficiently minimizes heat conduction. It's much better at reducing heat transfer than glass, which has a higher thermal conductivity.

Features of effective insulation materials include:
  • Low thermal conductivity: Minimize heat loss or gain.
  • Durability: Long-lasting effectiveness.
  • Cost-efficiency: Affordable with long-term savings on energy.
Choosing the right insulation material is crucial for trapping heat inside during cold months and keeping it out during warmer times, thereby creating stable indoor temperatures and reducing energy bills.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A baking dish is removed from a hot oven and placed on a cooling rack. As the dish cools down to \(35^{\circ} \mathrm{C}\) from \(175^{\circ} \mathrm{C}\), its net radiant power decreases to \(12.0 \mathrm{W}\). What was the net radiant power of the baking dish when it was first removed from the oven? Assume that the temperature in the kitchen remains at \(22^{\circ} \mathrm{C}\) as the dish cools down.

A person's body is covered with \(1.6 \mathrm{m}^{2}\) of wool clothing. The thickness of the wool is \(2.0 \times 10^{-3} \mathrm{m} .\) The temperature at the outside surface of the wool is \(11^{\circ} \mathrm{C},\) and the skin temperature is \(36^{\circ} \mathrm{C} .\) How much heat per second does the person lose due to conduction?

Liquid helium is stored at its boiling-point temperature of \(4.2 \mathrm{K}\) in a spherical container \((r=0.30 \mathrm{m})\). The container is a perfect blackbody radiator. The container is surrounded by a spherical shield whose temperature is \(77 \mathrm{K}\). A vacuum exists in the space between the container and the shield. The latent heat of vaporization for helium is \(2.1 \times 10^{4} \mathrm{J} / \mathrm{kg} .\) What mass of liquid helium boils away through a venting valve in one hour?

A solid sphere has a temperature of \(773 \mathrm{K}\). The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature?

A person is standing outdoors in the shade where the temperature is \(28^{\circ} \mathrm{C} .\) (a) What is the radiant energy absorbed per second by his head when it is covered with hair? The surface area of the hair (assumed to be flat) is \(160 \mathrm{cm}^{2}\) and its emissivity is \(0.85 .\) (b) What would be the radiant energy absorbed per second by the same person if he were bald and the emissivity of his head were \(0.65 ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.