What is the rate of heat conduction through a 30 m thickness of rhyolite lava?

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The discussion revolves around calculating the rate of heat conduction through 30 meters of rhyolite lava, with surface temperatures of 400°C and an interior temperature of 3100°C. The thermal conductivity of rhyolite is given as 0.84 W/m²°C. The correct calculation involves determining the temperature gradient by dividing the temperature difference (2700°C) by the thickness (30 m), leading to a gradient of 90°C/m. The final rate of heat conduction is confirmed to be 75.6 W/m², clarifying the confusion about whether to multiply or divide by the thickness. The resolution highlights the importance of correctly applying the formula for heat conduction.
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Homework Statement



A lava beds cool very slowly, remaining warm for years. Calculate the rate of heat conduction per square metre through a 30 m thickness of rhyolite, if the surface is 400 degrees celcius and its interior side is at 3100 degrees celcius. The thermal conductivity of rhyolite =0.84 Wm^-1 degrees celcius^-1.

Answer must be 75.6 W/m^2


Homework Equations



P=kA \left| \frac{dT}{dx}\right|

(The rate is P, k is thermal conductivity, dT/dx is temprature gradient)

The Attempt at a Solution



dT/dx = 3100-400 = 2700

0.84 \times 30 \times(2700) = 68040

Can anyone show me the problem? ...because the right answer is 75.6!
 
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Why are you multiplying by 30 instead of dividing? dT/dx is 2700/30.
 
ideasrule said:
Why are you multiplying by 30 instead of dividing? dT/dx is 2700/30.

Yes, but I need a number for area A!
 
Oh, never mind I get it. THANKS!
 

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