Solve Heat Conduction Problem: Oven Energy Use & Cost

[shaka23h]

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:tongue:

 

[OlderDan]

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:tongue:

Ya got to use da right numbas

 

[kyle8921]

Hi Jason,

Yes, your calculation for the energy used (Q) is correct. However, when converting from hours to seconds, make sure to use the conversion factor of 3600 seconds per hour, not 13968. This will give you a value of 3.88 hours = 13968 seconds.

Also, make sure to use the correct units for thermal conductivity (k) which is in units of J/(s m K). So, when you plug in your numbers, it should look like this:

Q = (0.0405 J/(s m C°)) (1.44 m^2) (160 C° - 38.6 C°) (3.88 hours * 3600 seconds/hour) / 0.0249 m

This will give you a final answer of 4412945.581 J, which is the energy used to operate the oven for 3.88 hours.

To calculate the cost, you need to convert this energy from joules to kilowatt-hours (kWh). Since 1 kWh = 3.6 x 10^6 J, the energy used in your problem is equal to:

Q = 4412945.581 J / (3.6 x 10^6 J/kWh) = 1.225 kWh

Now, to calculate the cost, you simply multiply this value by the cost per kWh, which is $0.10.

Cost = 1.225 kWh * $0.10/kWh = $0.1225

Therefore, the cost of operating the oven for 3.88 hours is $0.1225.

I hope this helps! Let me know if you have any other questions.