Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 8

Write the heat flux density \(300 \mathrm{~W} / \mathrm{m}^{2}\) with the units in the following forms: a) one that includes joules b) one that includes newtons c) one that includes kilograms

Short Answer

Expert verified
Answer: The heat flux density with Joules included is \(300\,\mathrm{J/(sm^2)}\), with Newtons included is \(300\,\mathrm{N/(sm)}\), and with kilograms included is \(300\,\mathrm{kg\, m^2/s^3}\).

Step by step solution

01

Part a: Joules

The given unit is \(\mathrm{W/m^2}\), and we want to rewrite it using Joules. We know that 1 Watt is equal to 1 Joule per second, or \(\mathrm{W = J/s}\). Therefore, we can rewrite the heat flux density using Joules as follows: \(300 \mathrm{~W/m^2} = 300 \mathrm{~(J/s)/m^2}\) So the heat flux density with Joules included is \(300\,\mathrm{J/(sm^2)}\).
02

Part b: Newtons

Now, we want to rewrite the heat flux density unit using Newtons. We know that 1 Joule is equal to 1 Newton meter, or \(\mathrm{J = N \cdot m}\). Therefore, we can first rewrite the heat flux density in terms of Newtons as follows: \(300 \mathrm{~(J/s)/m^2} = 300 \mathrm{~(N \cdot m/s) / m^2}\) Now, simplify the units: \(300 \mathrm{~(N \cdot m/s) / m^2} = 300\,\mathrm{N/(sm)}\) So the heat flux density with Newtons included is \(300\,\mathrm{N/(sm)}\).
03

Part c: Kilograms

Finally, we want to rewrite the heat flux density using kilograms. Since 1 Newton is equal to 1 kilogram-meter per second squared (from Newton's second law, \(F = m \cdot a\)), we can rewrite the heat flux density as follows: \(300 \mathrm{~(N/s)/m} = 300 \mathrm{ ~(kg \cdot m/s^2 \cdot s)/m}\) Now, simplify the units: \(300 \mathrm{ ~(kg \cdot m/s^2 \cdot s)/m} = 300 \mathrm{~ kg \cdot m^2/s^3}\) So the heat flux density with kilograms included is \(300\,\mathrm{kg\, m^2/s^3}\).

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.

Start your free trial

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

Select the angle that is approximately equal to a radian: a) 180 degrees b) 30 degrees c) 57 degrees d) 360 degrees

Write each of the following in a correct form without using any negative exponents: a) \(0.004562 \mathrm{kNm} \mathrm{m}^{-3} \mathrm{~K}^{-1}\) b) \(10^{-5} \mathrm{MNs}^{-1} \mathrm{~m}^{-3} \mathrm{~K}^{-1}\) c) \(0.000473001 \mathrm{mJs} \mathrm{s}^{-1} \mathrm{~m}^{-3} \mathrm{~kg}^{-1}\)

A jet plane is traveling at \(900 \mathrm{~km} / \mathrm{h}\) at an altitude where the temperature is such that the speed of sound is \(600 \mathrm{~km} / \mathrm{h}\). a) Determine the Mach number for the jet plane. b) How will the answer in part (a) be different if the speeds were given in miles per hour?

Write the viscosity \(\mu=0.001\) Pa.s using newtons and not pascals. Also, write the viscosity without using either newtons or pascals.

State both the primitive and the modern definitions of time. Which do you find the more reasonable?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Scientific Method Physics

Read Explanation

Linear Momentum

Read Explanation

Nuclear Physics

Read Explanation

Solid State 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.

Sign-up for free