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July 30, 2014

July 30, 2014

Posted by **Maryo** on Wednesday, March 24, 2010 at 1:36am.

- Physics -
**Maryo**, Wednesday, March 24, 2010 at 7:01amAnyone Can answer this Please

- Physics -
**drwls**, Wednesday, March 24, 2010 at 9:06amCalculate the heat loss rate through the walls, using

P = (surface area)*k*(110C)/0.02 m

where k is the thermal conducticity

Convert watts (J/s) to kilowatts. Multiply the number of kW by 6h to get the number of kW-hr.

That is the energy required for steady state ooperation for 6 hours. There will be an additional amount required to heat up the oven - probably a few percent more, depending upon the specific heat.

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