/*! 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 85 One end of a cylindrical iron ro... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 85

One end of a cylindrical iron rod of length \(1.00 \mathrm{m}\) and of radius \(1.30 \mathrm{cm}\) is placed in the blacksmith's fire and reaches a temperature of \(327^{\circ} \mathrm{C} .\) If the other end of the rod is being held in your hand \(\left(37^{\circ} \mathrm{C}\right),\) what is the rate of heat flow along the rod? The thermal conductivity of yal iron varies with temperature, but an average between the two temperatures is $67.5 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})$. (tutorial: conduction)

Short Answer

Expert verified
Answer: The rate of heat flow along the rod is approximately 10.46 W.

Step by step solution

01

Understand the formula for heat conduction

The formula for heat conduction in a cylindrical rod is given by: Q = -kA(螖T)/L where: Q = rate of heat flow (W) k = thermal conductivity (W/(m*K)) A = cross-sectional area of the rod (m虏) 螖T = temperature difference (K) L = length of the rod (m)
02

Calculate the cross-sectional area

Given the radius (r) of the cylindrical iron rod, we can calculate its cross-sectional area A using the formula: A = 蟺r虏 The radius is given as 1.3 cm which is equal to 0.013 m (in SI units): A = 蟺(0.013)虏 = 0.0005309 m虏
03

Calculate the temperature difference

We are given the two temperatures, T鈧� (fire) and T鈧� (hand), so we can calculate the temperature difference as follows: 螖T = T鈧� - T鈧� T鈧� = 327掳C = 327K (since there is no difference when converting from Celsius to Kelvin) T鈧� = 37掳C = 37K 螖T = 327 - 37 = 290 K
04

Calculate the rate of heat flow

Now that we have all the values, we can simply plug them into the heat conduction formula from Step 1: Q = -kA(螖T)/L k = 67.5 W/(m*K) A = 0.0005309 m虏 螖T = 290 K L = 1.00 m Q = -67.5 脳 0.0005309 脳 290 / 1.00 = -10.46 W (approx.) Note that the negative sign indicates that the heat flow is from the hot end to the cold end. The rate of heat flow along the rod is approximately 10.46 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

Consider the leaf of Problem \(70 .\) Assume that the top surface of the leaf absorbs \(70.0 \%\) of \(9.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2}\) of radiant energy, while the bottom surface absorbs all of the radiant energy incident on it due to its surroundings at \(25.0^{\circ} \mathrm{C} .\) (a) If the only method of heat loss for the leaf were thermal radiation, what would be the temperature of the leaf? (Assume that the leaf radiates like a blackbody.) (b) If the leaf is to remain at a temperature of \(25.0^{\circ} \mathrm{C}\) how much power per unit area must be lost by other methods such as transpiration (evaporative heat loss)?
It is a damp, chilly day in a New England seacoast town suffering from a power failure. To warm up the cold, clammy sheets, Jen decides to fill hot water bottles to tuck between the sheets at the foot of the beds. If she wishes to heat \(2.0 \mathrm{L}\) of water on the wood stove from $20.0^{\circ} \mathrm{C}\( to \)80.0^{\circ} \mathrm{C},$ how much heat must flow into the water?
A 75 -kg block of ice at \(0.0^{\circ} \mathrm{C}\) breaks off from a glacier, slides along the frictionless ice to the ground from a height of $2.43 \mathrm{m},$ and then slides along a horizontal surface consisting of gravel and dirt. Find how much of the mass of the ice is melted by the friction with the rough surface, assuming \(75 \%\) of the internal energy generated is used to heat the ice.
A cylinder contains 250 L of hydrogen gas \(\left(\mathrm{H}_{2}\right)\) at \(0.0^{\circ} \mathrm{C}\) and a pressure of 10.0 atm. How much energy is required to raise the temperature of this gas to \(25.0^{\circ} \mathrm{C} ?\)
How much heat is required to change \(1.0 \mathrm{kg}\) of ice, originally at \(-20.0^{\circ} \mathrm{C},\) into steam at \(110.0^{\circ} \mathrm{C} ?\) Assume 1.0 atm of pressure.
See all solutions

Recommended explanations on Physics Textbooks

Turning Points in Physics

Read Explanation

Engineering Physics

Read Explanation

Measurements

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Astrophysics

Read Explanation

Energy 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.