Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider two houses that are identical except that the walls are built using bricks in one house and wood in the other. If the walls of the brick house are twice as thick, which house do you think will be more energy efficient?

Short Answer

Expert verified
Answer: The wood house is more energy efficient due to its lower heat transfer rate compared to the brick house.

Step by step solution

01

Define heat transfer rate equation

We will use the heat transfer equation to calculate the rate at which heat is transferred through the walls of both houses. The heat transfer rate through a surface is given by: Q = U * A * ∆T Where Q is the heat transfer rate, U is the overall heat transfer coefficient, A is the surface area in contact, and ∆T is the temperature difference across the surface. We need to find the heat transfer coefficient (U) for both brick and wood to compare their energy efficiency.
02

Calculate heat transfer coefficients for brick and wood

We will use the thermal conductivity (k) values of brick and wood to find the heat transfer coefficients for both materials. Thermal conductivity of brick: k_b = 0.6 W/(m*K) Thermal conductivity of wood: k_w = 0.15 W/(m*K) Thickness of brick wall: d_b = 2*d_w (since the brick wall is twice as thick) Thickness of wood wall: d_w The heat transfer coefficient U is determined by: U = k / d For brick: U_b = k_b / d_b For wood: U_w = k_w / d_w However, since we don't have the values of both thicknesses, we can analyze the ratio of heat transfer coefficients to determine the relative energy efficiency of both houses.
03

Compare the heat transfer coefficients

By analyzing the ratio of heat transfer coefficients, we can determine which house is more energy efficient. U_b / U_w = (k_b / d_b) / (k_w / d_w) Substituting d_b = 2*d_w U_b / U_w = (k_b / (2*d_w)) / (k_w / d_w) U_b / U_w = (0.6 / (2*d_w)) / (0.15 / d_w) After simplification, we find: U_b / U_w = 2
04

Determine the more energy efficient house

Since the ratio of U_b to U_w is 2, this means that the heat transfer rate in the brick house is twice as high as that in the wood house. Thus, the wood house has a lower heat transfer rate, making it more energy efficient. In conclusion, the wood house will be more energy efficient due to its lower heat transfer rate compared to the brick house.

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 Vaia!

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 10-cm-high and 20-cm-wide circuit board houses on its surface 100 closely spaced chips, each generating heat at a rate of \(0.08 \mathrm{~W}\) and transferring it by convection and radiation to the surrounding medium at \(40^{\circ} \mathrm{C}\). Heat transfer from the back surface of the board is negligible. If the combined convection and radiation heat transfer coefficient on the surface of the board is $22 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, the average surface temperature of the chips is (a) \(72.4^{\circ} \mathrm{C}\) (b) \(66.5^{\circ} \mathrm{C}\) (c) \(40.4^{\circ} \mathrm{C}\) (d) \(58.2^{\circ} \mathrm{C}\) (e) \(49.1^{\circ} \mathrm{C}\)

A cold bottled drink ( $\left.m=2.5 \mathrm{~kg}, c_{p}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( at \)5^{\circ} \mathrm{C}$ is left on a table in a room. The average temperature of the drink is observed to rise to \(15^{\circ} \mathrm{C}\) in \(30 \mathrm{~min}\). The average rate of heat transfer to the drink is (a) \(23 \mathrm{~W}\) (b) \(29 \mathrm{~W}\) (c) \(58 \mathrm{~W}\) (d) \(88 \mathrm{~W}\) (e) \(122 \mathrm{~W}\)

The north wall of an electrically heated home is \(20 \mathrm{ft}\) long, $10 \mathrm{ft}\( high, and \)1 \mathrm{ft}$ thick and is made of brick whose thermal conductivity is $k=0.42 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}$. On a certain winter night, the temperatures of the inner and the outer surfaces of the wall are measured to be at about \(62^{\circ} \mathrm{F}\) and \(25^{\circ} \mathrm{F}\), respectively, for a period of \(8 \mathrm{~h}\). Determine \((a)\) the rate of heat loss through the wall that night and \((b)\) the cost of that heat loss to the homeowner if the cost of electricity is \(\$ 0.07 / \mathrm{kWh}\).

Consider a person standing in a room at \(23^{\circ} \mathrm{C}\). Determine the total rate of heat transfer from this person if the exposed surface area and the skin temperature of the person are \(1.7 \mathrm{~m}^{2}\) and $32^{\circ} \mathrm{C}\(, respectively, and the convection heat transfer coefficient is \)5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Take the emissivity of the skin and the clothes to be \(0.9\), and assume the temperature of the inner surfaces of the room to be the same as the air temperature.

A series of experiments were conducted by passing \(40^{\circ} \mathrm{C}\) air over a long 25 -mm-diameter cylinder with an embedded electrical heater. The objective of these experiments was to determine the power per unit length required \((W / L)\) to maintain the surface temperature of the cylinder at \(300^{\circ} \mathrm{C}\) for different air velocities \((V)\). The results of these experiments are given in the following table: $$ \begin{array}{lccccc} \hline V(\mathrm{~m} / \mathrm{s}) & 1 & 2 & 4 & 8 & 12 \\ W / L(\mathrm{~W} / \mathrm{m}) & 450 & 658 & 983 & 1507 & 1963 \\ \hline \end{array} $$ (a) Assuming a uniform temperature over the cylinder, negligible radiation between the cylinder surface and surroundings, and steady-state conditions, determine the convection heat transfer coefficient \((h)\) for each velocity \((V)\). Plot the results in terms of $h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\( vs. \)V(\mathrm{~m} / \mathrm{s})$. Provide a computer- generated graph for the display of your results, and tabulate the data used for the graph. (b) Assume that the heat transfer coefficient and velocity can be expressed in the form \(h=C V^{n}\). Determine the values of the constants \(C\) and \(n\) from the results of part (a) by plotting \(h\) vs. \(V\) on log-log coordinates and choosing a \(C\) value that assures a match at \(V=1 \mathrm{~m} / \mathrm{s}\) and then varying \(n\) to get the best fit.

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.

Sign-up for free