/*! 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 67 A black wood stove has a surface... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 67

A black wood stove has a surface area of \(1.20 \mathrm{m}^{2}\) and a surface temperature of \(175^{\circ} \mathrm{C} .\) What is the net rate at which heat is radiated into the room? The room temperature is \(20^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Answer: The net rate at which heat is radiated into the room is approximately 920.83 W.

Step by step solution

01

Convert temperatures to Kelvin

To use the Stefan-Boltzmann Law, we need to convert the given temperatures to Kelvin. The conversion formula is K = C + 273.15. Stove temperature in Kelvin: \(T_s = 175 + 273.15 = 448.15 \mathrm{K}\) Room temperature in Kelvin: \(T_r = 20 + 273.15 = 293.15 \mathrm{K}\)
02

Write the Stefan-Boltzmann Law for power emitted and absorbed

The Stefan-Boltzmann Law states that the power radiated per unit area by a black body is directly proportional to the fourth power of its absolute temperature. \(P = \sigma A T^4\), where \(P\) = radiated power, \(\sigma\) = Stefan-Boltzmann constant \((5.67 \times 10^{-8}\) W m\(^{-2}\) K\(^{-4})\), \(A\) = surface area, and \(T\) = absolute temperature.
03

Calculate the power emitted by the stove

Using the Stefan-Boltzmann Law, we can find the power emitted by the stove. \(P_s = \sigma A_s {T_s}^4\) \(P_s = (5.67 \times 10^{-8}\) W m\(^{-2}\) K\(^{-4})\) 脳 \(1.20 \mathrm{m^2}\) 脳 \((448.15 \mathrm{K})^4\) \(P_s \approx 1241.99 \mathrm{W}\)
04

Calculate the power absorbed by the room

Similarly, we can find the power absorbed by the room. \(P_r = \sigma A_s {T_r}^4\) \(P_r = (5.67 \times 10^{-8}\) W m\(^{-2}\) K\(^{-4})\) 脳 \(1.20 \mathrm{m^2}\) 脳 \((293.15 \mathrm{K})^4\) \(P_r \approx 321.16 \mathrm{W}\)
05

Find the net rate of heat radiation

The net rate of heat radiation is the difference between the power emitted by the stove and the power absorbed by the room. Net rate of heat radiation = \(P_s - P_r\) Net rate of heat radiation = \(1241.99 \mathrm{W} - 321.16 \mathrm{W}\) Net rate of heat radiation \(\approx 920.83 \mathrm{W}\) Thus, the net rate at which heat is radiated into the room is approximately \(920.83 \mathrm{W}\).

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 91影视!

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

In a physics lab, a student accidentally drops a \(25.0-\mathrm{g}\) brass washer into an open dewar of liquid nitrogen at 77.2 K. How much liquid nitrogen boils away as the washer cools from \(293 \mathrm{K}\) to $77.2 \mathrm{K} ?\( The latent heat of vaporization for nitrogen is \)199.1 \mathrm{kJ} / \mathrm{kg}$.
A \(0.400-\mathrm{kg}\) aluminum teakettle contains \(2.00 \mathrm{kg}\) of water at \(15.0^{\circ} \mathrm{C} .\) How much heat is required to raise the temperature of the water (and kettle) to \(100.0^{\circ} \mathrm{C} ?\)
At a tea party, a coffeepot and a teapot are placed on the serving table. The coffeepot is a shiny silver-plated pot with emissivity of \(0.12 ;\) the teapot is ceramic and has an emissivity of \(0.65 .\) Both pots hold \(1.00 \mathrm{L}\) of liquid at \(98^{\circ} \mathrm{C}\) when the party begins. If the room temperature is at \(25^{\circ} \mathrm{C},\) what is the rate of radiative heat loss from the two pots? [Hint: To find the surface area, approximate the pots with cubes of similar volume.]
The amount of heat generated during the contraction of muscle in an amphibian's leg is given by $$Q=0.544 \mathrm{mJ}+(1.46 \mathrm{mJ} / \mathrm{cm}) \Delta x$$ where \(\Delta x\) is the length shortened. If a muscle of length \(3.0 \mathrm{cm}\) and mass \(0.10 \mathrm{g}\) is shortened by $1.5 \mathrm{cm}$ during a contraction, what is the temperature rise? Assume that the specific heat of muscle is $4.186 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)$.
(a) What thickness of cork would have the same R-factor as a 1.0 -cm thick stagnant air pocket? (b) What thickness of tin would be required for the same R-factor?
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Dynamics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Measurements

Read Explanation

Magnetism

Read Explanation

Fundamentals of Physics

Read Explanation
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.