Rate of Energy Transferred, Aluminum Pot

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SUMMARY

The discussion centers on calculating the rate of energy transfer through the bottom of a circular aluminum pot using the equation P=KA(TH-TC)/L. The thermal conductivity (K) of aluminum is noted as 220 k (J/s⋅m⋅ºC), with the pot's radius of 8 cm and thickness of 1 cm. The correct cross-sectional area (A) is derived from the radius, and the temperature difference (TH-TC) is calculated as 100 degrees Celsius. The final calculated power output is approximately 44230W, which aligns closely with option d (4.79X10^4W).

PREREQUISITES
  • Understanding of thermal conductivity and its units (J/s⋅m⋅ºC)
  • Familiarity with the formula for heat transfer (P=KA(TH-TC)/L)
  • Basic geometry to calculate the cross-sectional area of a circle
  • Knowledge of temperature scales and conversions
NEXT STEPS
  • Research the properties of aluminum and its thermal conductivity in various conditions
  • Learn how to accurately calculate the cross-sectional area of circular objects
  • Explore the implications of heat transfer in cooking and material science
  • Investigate the reliability of sources for physical constants in scientific calculations
USEFUL FOR

Students in physics or engineering fields, educators teaching thermodynamics, and anyone interested in the practical applications of heat transfer principles.

DracoMalfoy
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Homework Statement


The bottom of a circular, aluminum pot has a radius of 8cm and a thickness of 1cm. The temperature of the stove top is 120 degrees celsius and the temperature of the pot initially is 20 degrees celsius. What is the rate at which energy is transferred through the bottom of the pot?

a. 300W

b. 2589W

c. 2.67X10^5W

d. 4.79X10^4W

e. 7.89W

Homework Equations



P=KA(TH-TC)/L

The Attempt at a Solution


[/B]
Thermal Conductivity(K) of Aluminum: 220 k (J/s⋅m⋅ºC)
A(Cross sectional area the energy goes through): .08m??
(Th-Tc): (120-20degreeC)
L(Distance energy travels): .01m?

Im not sure if this is the right equation to use, or if A or L is correct. I am not getting the answer this way.
 
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DracoMalfoy said:
if this is the right equation to use, or if A or L is correct
Right equation, right L, but the 8cm is a radius, not an area.
 
haruspex said:
Right equation, right L, but the 8cm is a radius, not an area.
i understand. i got the answer. d i think
 
DracoMalfoy said:
i understand. i got the answer. d i think
Yes, that's the nearest to what I get ... 44230W ... but worryingly different.
 
haruspex said:
Yes, that's the nearest to what I get ... 44230W ... but worryingly different.
I've noticed that the OP has several posts where the source of relevant constants is a tad dubious and differ from other more reliable sources by a fairly substantial percent. I'm not sure if these sources are mandated by the course or randomly chosen by the OP, but either way it's a bit disconcerting.
 
gneill said:
I've noticed that the OP has several posts where the source of relevant constants is a tad dubious and differ from other more reliable sources by a fairly substantial percent. I'm not sure if these sources are mandated by the course or randomly chosen by the OP, but either way it's a bit disconcerting.
I wondered about the conductivity figure, but if anything it seems a little high.
 

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