Thermal Conduction - Finding Temperature of Interface Between 2 Slabs

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SUMMARY

The discussion focuses on calculating the temperature at the interface between two slabs: steel and copper, with thermal conductivities of 50 W/m·K and 385 W/m·K, respectively. The left side of the steel slab is at 427°C, while the right side of the copper slab is at 77°C. The slabs have areas of 90 cm² and lengths of 20 cm for steel and 30 cm for copper. The solution involves applying the concept of thermal resistance in series to determine the interface temperature and the rate of heat transfer.

PREREQUISITES
  • Understanding of thermal conductivity and its units (W/m·K)
  • Knowledge of heat transfer equations, specifically Fourier's law
  • Familiarity with the concept of thermal resistance in series
  • Basic algebra for solving equations
NEXT STEPS
  • Study Fourier's law of heat conduction in detail
  • Learn about thermal resistance calculations for composite materials
  • Explore the concept of steady-state heat transfer
  • Investigate practical applications of thermal conductivity in engineering
USEFUL FOR

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

MadmanMurray
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Homework Statement


Two well-insulated slabs, one of steel the other of copper, are in close contact as illustrated.

Find the temperature at the interface between the two slabs & also the rate at which heat is transferred across the slabs.

Thermal conductivities of steel and copper are 50WmK and 385WmK

Left side of steel slab has an area of 90cm2 and temperature of 427C. The steel slab has a length of 20cm.

Right side of copper slab has area of 90cm2 and temperature of 77C. Length of slab is 30cm



Homework Equations





The Attempt at a Solution


I'm completely stuck. I know a formula to find the rate of heat transfer through the slabs but I don't have a clue how to find the temperature at the interface of these 2 slabs.
 
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Just think of them as resistors in series = a potential divider.
 

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