Heat Flowing Through A Sectioned Rod HELP

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SUMMARY

The discussion focuses on calculating the heat transfer through a sectioned rod made of copper and steel, with specific thermal conductivities of 385 W/m·K for copper and 50.2 W/m·K for steel. The rod is insulated, with one end in boiling water and the other in an ice-water mixture, resulting in a junction temperature of 65.0°C. The correct formula for heat transfer, H = k * A * (ΔT) / L, is applied, and the area must be converted from cm² to m² for accurate calculations. The final heat transfer rate calculated for the copper section is 21.56 W, which was initially miscalculated due to incorrect area units.

PREREQUISITES
  • Understanding of thermal conductivity and heat transfer principles
  • Familiarity with the formula H = k * A * (ΔT) / L
  • Knowledge of unit conversions, specifically area from cm² to m²
  • Basic concepts of steady-state heat transfer in composite materials
NEXT STEPS
  • Review the derivation and applications of the heat transfer equation H = k * A * (ΔT) / L
  • Learn about thermal conductivity values for various materials and their implications in engineering
  • Study the effects of insulation on heat transfer in practical applications
  • Explore numerical methods for solving heat transfer problems in composite materials
USEFUL FOR

Students studying thermodynamics, engineers working with thermal systems, and anyone involved in heat transfer analysis in materials science.

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Homework Statement



A long rod, insulated to prevent heat loss along its sides, is in perfect thermal contact with boiling water (at atmospheric pressure) at one end and with an ice-water mixture at the other. The rod consists of a 1.00 section of copper (with one end in the boiling water) joined end-to-end to a length of steel (with one end in the ice water). Both sections of the rod have cross-sectional areas of 4.00 . The temperature of the copper-steel junction is 65.0 after a steady state has been reached.
How much heat per second flows from the boiling water to the ice-water mixture?
Express your answer in watts

Homework Equations



H=kAdT/L
kcopper=385
ksteel=50.2

The Attempt at a Solution



Hcopper=Hsteel
Hcopper=385*0.04^2*35
= 21.56W
There's something wrong with this answer.. Can someone please help me?
Thank you!
 
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Hi tizzful,

In your expression you have squared 0.04 (which is (area/length) in meters); since they gave the area I don't believe that number should be squared.
 
Hello, yes I realized afterwards that its in cm squared and I need it in meters squared and therefore had to multiply it by 10^-4. Thank you
 

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