Solve Heat Conduction Problem: Oven Energy Use & Cost

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SUMMARY

The discussion focuses on calculating the energy consumption and cost of operating an electric oven at a temperature of 160 °C with an outer kitchen temperature of 38.6 °C. The formula used is Q = (k A ΔT t) / L, where k is the thermal conductivity (0.0405 J/(s m °C)), A is the surface area (1.44 m²), ΔT is the temperature difference (121.4 °C), t is the time in seconds (13968 seconds), and L is the insulation thickness (0.0249 m). The calculated energy expenditure is 4,412,945.581 J, but the user encountered an error when verifying this value with a web-based solution.

PREREQUISITES
  • Understanding of heat conduction principles
  • Familiarity with the formula Q = (k A ΔT t) / L
  • Basic knowledge of unit conversions (hours to seconds)
  • Experience with energy cost calculations
NEXT STEPS
  • Review the concept of thermal conductivity and its units
  • Learn about energy conversion from joules to kilowatt-hours
  • Explore common errors in heat conduction calculations
  • Investigate insulation materials and their thermal properties
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Engineers, physicists, and anyone involved in thermal management or energy efficiency in appliances will benefit from this discussion.

shaka23h
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The temperature in an electric oven is 160 °C. The temperature at the outer surface in the kitchen is 38.6 °C. The oven (surface area = 1.44 m2) is insulated with material that has a thickness of 0.0249 m and a thermal conductivity of 0.0405 J/(s m C°). (a) How much energy is used to operate the oven for 3.88 hours? (b) At a price of $0.10 per kilowatt-hour for electrical energy, what is the cost (in dollars) of operating the oven?


OK I know to use the Conduction of Heat Through A material on this problem

Q = (k A Delta T) t/ L

After plugging in my values Q = [.045 J/ (s.m.C (degrees)] (1.44m ^2) (160 degrees - 38.6 degrees)(13968 sec) (hours converted to seconds) / 0.0249 m


I get my answer to be 4412945.581 J? This should be my energy value right? The Q in this case is my energy expenditure?


I plugged this answer in for the webbased solution but it was incorrect.

Thanks for ur help

Jason:-p
 
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shaka23h said:
The temperature in an electric oven is 160 °C. The temperature at the outer surface in the kitchen is 38.6 °C. The oven (surface area = 1.44 m2) is insulated with material that has a thickness of 0.0249 m and a thermal conductivity of 0.0405 J/(s m C°). (a) How much energy is used to operate the oven for 3.88 hours? (b) At a price of $0.10 per kilowatt-hour for electrical energy, what is the cost (in dollars) of operating the oven?


OK I know to use the Conduction of Heat Through A material on this problem

Q = (k A Delta T) t/ L

After plugging in my values Q = [.045 J/ (s.m.C (degrees)] (1.44m ^2) (160 degrees - 38.6 degrees)(13968 sec) (hours converted to seconds) / 0.0249 m


I get my answer to be 4412945.581 J? This should be my energy value right? The Q in this case is my energy expenditure?


I plugged this answer in for the webbased solution but it was incorrect.

Thanks for ur help

Jason:-p
Ya got to use da right numbas
 

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