How to Calculate Heat Transfer Rate Through a Furnace Wall?

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SUMMARY

The discussion centers on calculating the heat transfer rate through a furnace wall with specific parameters: an area of 3 m², a thickness of 250 mm, and a thermal conductivity of 1.5 W m-1 K-1. The key equation used is Q/t = KxAx(T2-T1)/L. Participants debated the interpretation of a temperature difference of 40 K, with suggestions that it may be a typo, as typical furnace temperatures are significantly higher, often around 350 K or more. The consensus is that assuming a delta T of 40 K is necessary to proceed with the calculation, despite concerns about its validity.

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Homework Statement



a furnace wall has an area of 3m^2. a thickness of 250mm and a thermal conductivity of 1.5 W m^-1 K^-1. calculate the rate of heat transfer through it when the temperature is 40K

Homework Equations



Q/t=KxAx(T2-T1)/L

The Attempt at a Solution



This isn't homework as such, it was given in a list of questions aimed at helping with revision for my module exam. I've done the other 11 questions but this one is doing my head in. the furthest I've gotten to is (Q/18t)-40=T2 but without knowing T2 or the transfer rate is it impossible to progress to an answer?

please help. don't want to be revising something incorrectly for the exam
 
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40K must be the temperature difference (delta T) it surely isn't -233 C for a furnace.
 
i was talking to my dad about it earlier today too. but that's the exact question, haven't re worded it at all. a furnace implies very hot temperatures, and the 40k must have to be a difference. but i wanted to see if i was overlooking something so posted on here.

is there anyway to get any further on this question with the lack of information?
 
I don't see any other way to solve this problem rather than to assume a delta T of 40 K. But actually that value seems to low. Furnace temperatures get very high, I would have expected a delta T of about 350 K assuming an exterior surface temperature of about 40 C (100 F) and an inside wall temperature of 380 C (800 F). Might be a typo.
 
right cheers for that then. i know how to do it providing that is is delta 40, i also know how to do it given 2 temperatures so cheers for the help. was annoying me greatly

much appreciated, ash
 

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