Calculating Heat Flow Through a Block of Material

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SUMMARY

The discussion centers on calculating the rate of heat flow through a block of wood material with a cross-sectional area of 15 cm² and a length of 8 cm, given a temperature difference of 30 degrees Celsius. Using the formula P = KA(ΔT)/L, where K is the thermal conductivity (0.08 Js⁻¹m⁻¹C⁻¹), the calculated heat flow rate is 4.5 Watts. This result conflicts with the book's answer of 5.6 Watts, prompting a review of unit conversions and calculations. The discrepancy suggests a potential error in the textbook or a misunderstanding of the problem parameters.

PREREQUISITES
  • Understanding of thermal conductivity and its units
  • Familiarity with the formula for heat transfer (P = KA(ΔT)/L)
  • Basic unit conversion skills (cm² to m², etc.)
  • Knowledge of heat flow concepts in physics
NEXT STEPS
  • Review unit conversion techniques for area and length in thermal calculations
  • Study the principles of heat transfer in different materials
  • Learn about common errors in thermal conductivity calculations
  • Explore advanced topics in thermodynamics related to heat flow
USEFUL FOR

Students studying physics, particularly those focusing on thermodynamics, as well as educators and professionals involved in material science and engineering applications related to heat transfer.

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Homework Statement


A block of material with a cross sectional area of 15cm^2 and length of 8cm is at hand. A temperature difference of 30degrees is established and maintained across the block. Find the rate of heat flow through it if the material is wood (K = 0.08Js-1m-1C-1)


Homework Equations



P = KA(change in T)/L

The Attempt at a Solution


P is watts. I converted area and length into meters.

Everything is given, all i need to do is plug in the values. I did that and I get 4.5 Watts. The answer in the back of the book is 5.6W.

Can someone do the calculation and confirm it with me? Either it's an error in the book or I'm missing something.

THANKS! :smile:
 
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I plugged the values exactly as given into my calculator, and I get an answer that differs from yours by a factor of 1000. I'd suggest double-checking your unit conversions.

I really don't know where the 5.6W in the back of the book might have come from, unless there's something more to the problem - but I can't think of what that might be.
 

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