Calculate Change in Entropy of Factory Furnace per sq m/sec

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SUMMARY

The discussion focuses on calculating the change in entropy of a factory furnace per square meter per second, given specific thermal conditions. The furnace operates at 175 degrees Celsius, while the outer surface of the insulation is at 42 degrees Celsius. The thermal conductivity of the insulation is 5.50×10−2 W/(m * K), and heat escapes at a rate of 125 W per square meter. The entropy change is calculated using the formula ΔS = Q/T, where Q is the heat transfer and T is the temperature in Kelvin.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically entropy.
  • Familiarity with heat transfer concepts, including thermal conductivity.
  • Knowledge of the relationship between heat (H), heat transfer (Q), and time (Δt).
  • Basic proficiency in converting temperatures from Celsius to Kelvin.
NEXT STEPS
  • Study the derivation and application of the entropy formula ΔS = Q/T.
  • Learn about heat transfer mechanisms in insulating materials.
  • Explore the relationship between thermal conductivity and heat loss in systems.
  • Investigate the implications of temperature gradients on entropy changes in thermodynamic systems.
USEFUL FOR

Students in thermodynamics, engineers working with thermal systems, and anyone interested in understanding entropy changes in heat transfer applications.

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Homework Statement


A large factory furnace maintained at 175 degrees C at its outer surface is wrapped in an insulating blanket of thermal conductivity 5.50×10−2 W/(m * K) which is thick enough that the outer surface of the insulation is at 42.0 degrees C while heat escapes from the furnace at a steady rate of 125 W for each square meter of surface area.

By how much does each square meter of the furnace change the entropy of the factory every second?

Homework Equations


\DeltaS = Q/T


The Attempt at a Solution


I know:
k = 5.50*10-2 W/(m*K)
H = 125 W
TH = 175 C = 448K
TC = 42 C = 315K

H = \DeltaQ/\Deltat

I know there is some type of relationship between H and Q, I just can't figure out what it is. Or how they work together.
 
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H =\DeltaQ/\Deltat = kA((TH - TC)/L)

L = 5.85 cm
But, does this help me in anyway?

S = Q/T
Am I supposed to figure out the entropy inside the furnace (Q/175 deg C) and add/subtract it from the entropy on the outside (Q/42 deg C)?
 

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