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PhysicsCollege Physics11th Edition

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Chapter 11, Problem 45P
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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

Solutions

Chapter
Section
Problem 1CQ
Problem 2CQ
Problem 3CQ
Problem 4CQ
Problem 5CQ
Problem 6CQ
Problem 7CQ
Problem 8CQ
Problem 9CQ
Problem 10CQ
Problem 11CQ
Problem 12CQ
Problem 13CQ
Problem 14CQ
Problem 15CQ
Problem 16CQ
Problem 1P
Problem 2P
Problem 3P
Problem 4P
Problem 5P
Problem 6P
Problem 7P
Problem 8P
Problem 9P
Problem 10P
Problem 11P
Problem 12P
Problem 13P
Problem 14P
Problem 15P
Problem 16P
Problem 17P
Problem 18P
Problem 19P
Problem 20P
Problem 21P
Problem 22P
Problem 23P
Problem 24P
Problem 25P
Problem 26P
Problem 27P
Problem 28P
Problem 29P
Problem 30P
Problem 31P
Problem 32P
Problem 33P
Problem 34P
Problem 35P
Problem 36P
Problem 37P
Problem 38P
Problem 39P
Problem 40P
Problem 41P
Problem 42P
Problem 43P
Problem 44P
Problem 45P
Problem 46P
Problem 47P
Problem 48P
Problem 49P
Problem 50P
Problem 51P
Problem 52P
Problem 53P
Problem 54P
Problem 55AP
Problem 56AP
Problem 57AP
Problem 58AP
Problem 59AP
Problem 60AP
Problem 61AP
Problem 62AP
Problem 63AP
Problem 64AP
Problem 65AP
Problem 66AP
Problem 67AP
Problem 68AP
Problem 69AP
Problem 70AP
Problem 71AP
Problem 72AP
Problem 73AP
Problem 74AP
Problem 75AP
BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

A steam pipe is covered with 1.50-cm-thick insulating material of thermal conductivity 0.200 cal/cm · °C · s. How much energy is lost every second when the steam is at 200.°C and the surrounding air is at 20.0°C? The pipe has a circumference of 800. cm and a length of 50.0 m. Neglect losses through the ends of the pipe.

To determine
The amount of energy lost every second.

Explanation

Given info: The thickness of the pipe is 1.50 cm, thermal conductivity of insulating material is 0.200 cal/cm⋅°C⋅s , temperature of the steam is 200°C , temperature of the surrounding air is at 20.0°C , the circumference of the pipe is 800 cm and its length is 50.0 m.

Formula to calculate the rate at which energy is transferred is,

P=kA(Th−Tc)L

  • P is the rate of energy transfer,
  • k is the thermal conductivity of insulating material,
  • Th is temperature of the steam,
  • Tc is temperature of the surrounding,
  • L is the thickness of the pipe,
  • A is the surface area of the pipe,

Formula to calculate the surface area of the pipe is,

A=C⋅l

  • C is the circumference of the pipe,
  • l is the length of the pipe,

Use C⋅l for l in P=kA(Th−Tc)L to rewrite P .

P=k(C⋅l)(Th−Tc)L

Substitute 0.200 cal/cm⋅°C⋅s for k, 1

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Sect-11.2 P-11.1QQSect-11.4 P-11.2QQSect-11.5 P-11.3QQSect-11.5 P-11.4QQSect-11.5 P-11.5QQCh-11 P-1CQCh-11 P-2CQCh-11 P-3CQCh-11 P-4CQCh-11 P-5CQ
Ch-11 P-6CQCh-11 P-7CQCh-11 P-8CQCh-11 P-9CQCh-11 P-10CQCh-11 P-11CQCh-11 P-12CQCh-11 P-13CQCh-11 P-14CQCh-11 P-15CQCh-11 P-16CQCh-11 P-1PCh-11 P-2PCh-11 P-3PCh-11 P-4PCh-11 P-5PCh-11 P-6PCh-11 P-7PCh-11 P-8PCh-11 P-9PCh-11 P-10PCh-11 P-11PCh-11 P-12PCh-11 P-13PCh-11 P-14PCh-11 P-15PCh-11 P-16PCh-11 P-17PCh-11 P-18PCh-11 P-19PCh-11 P-20PCh-11 P-21PCh-11 P-22PCh-11 P-23PCh-11 P-24PCh-11 P-25PCh-11 P-26PCh-11 P-27PCh-11 P-28PCh-11 P-29PCh-11 P-30PCh-11 P-31PCh-11 P-32PCh-11 P-33PCh-11 P-34PCh-11 P-35PCh-11 P-36PCh-11 P-37PCh-11 P-38PCh-11 P-39PCh-11 P-40PCh-11 P-41PCh-11 P-42PCh-11 P-43PCh-11 P-44PCh-11 P-45PCh-11 P-46PCh-11 P-47PCh-11 P-48PCh-11 P-49PCh-11 P-50PCh-11 P-51PCh-11 P-52PCh-11 P-53PCh-11 P-54PCh-11 P-55APCh-11 P-56APCh-11 P-57APCh-11 P-58APCh-11 P-59APCh-11 P-60APCh-11 P-61APCh-11 P-62APCh-11 P-63APCh-11 P-64APCh-11 P-65APCh-11 P-66APCh-11 P-67APCh-11 P-68APCh-11 P-69APCh-11 P-70APCh-11 P-71APCh-11 P-72APCh-11 P-73APCh-11 P-74APCh-11 P-75AP

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