# Solve Heat Conduction Problem: Oven Energy Use & Cost

• shaka23h
In summary: So, ta answer part (a), ya gotta use 1 joule = 0.00027777777 kilowatt-hours and ta answer part (b), ya gotta multiply da answer by 0.10 to get da cost in dollars. In summary, the energy used to operate the oven for 3.88 hours is 4412945.581 J and the cost of operating the oven is $12.26. 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

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 Ya got to use da right numbas 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.

## What is heat conduction and how does it relate to oven energy use?

Heat conduction is the transfer of heat within a material or between different materials that are in direct contact, such as the walls of an oven and the food being cooked. In an oven, heat conduction is responsible for transferring the heat from the heating elements to the food, which results in the energy use and cost of operating the oven.

## What factors affect the heat conduction in an oven?

The main factors that affect heat conduction in an oven include the type of material used for the oven walls, the thickness of the walls, the type and efficiency of the heating elements, and the type and amount of food being cooked.

## How can I reduce the energy use and cost of my oven?

To reduce the energy use and cost of your oven, you can try using materials with better insulation properties for the oven walls, ensuring proper sealing of the oven door, using energy-efficient heating elements, and cooking multiple items at once to make the most of the oven's energy consumption.

## How can I calculate the heat conduction and energy use of my oven?

The heat conduction and energy use of an oven can be calculated by measuring the temperature of the oven walls, the heating elements, and the food being cooked and using the principles of heat transfer to determine the heat conduction and energy consumption.

## What are some common misconceptions about heat conduction and oven energy use?

Some common misconceptions about heat conduction and oven energy use include thinking that leaving the oven door open will save energy, using aluminum foil as a heat barrier will reduce energy consumption, and cooking at a higher temperature will cook food faster and save energy. However, these actions can actually increase energy use and cost in the long run.

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