How much energy is used to operate the oven for 7 hours?

In summary, the temperature in an electric oven is 158°C and the temperature at the outer surface in the kitchen is 51°C. The oven, with a surface area of 1.8 m^2, is insulated with material that has a thickness of 0.023 m and a thermal conductivity of 0.045 J/(s·m·°C). To calculate the energy used to operate the oven for 7 hours, the power through the oven wall is multiplied by the time and divided by the thickness. The correct units for k are W/mK, and the correct answer is 9,500,000 J.
  • #1
keemosabi
109
0

Homework Statement


The temperature in an electric oven is 158°C. The temperature at the outer surface in the kitchen is 51°C. The oven (surface area = 1.8 m2) is insulated with material that has a thickness of 0.023 m and a thermal conductivity of 0.045 J/(s·m·°C).

(a) How much energy is used to operate the oven for 7 hours?


Homework Equations


Q = (k A deltaT time)/L


The Attempt at a Solution


k = .045
A = 1.8
deltaT = 107
time = 7
L = .023

I got 2637.78 but it says that's wrong. What did I do wrong?
 
Physics news on Phys.org
  • #2
Check units. I suspect a mistake in the units of k = 0.045 J/(s·m·°C). Perhaps m squared?
Also your 7 hours should be expressed in terms of the time unit in k so that the time cancels out in the calculation.
 
  • #3
Delphi51 said:
Check units. I suspect a mistake in the units of k = 0.045 J/(s·m·°C). Perhaps m squared?
Also your 7 hours should be expressed in terms of the time unit in k so that the time cancels out in the calculation.
The units of the k is correct. I double checked it. What do you mean by the 7 hours?
 
  • #4
keemosabi said:
What do you mean by the 7 hours?
You can't multiply hours by J/(s·m·°C).
 
  • #5
mgb_phys said:
You can't multiply hours by J/(s·m·°C).

So convert it to seconds? I did that and still got the wrong answer.
 
  • #6
Conductivity has units W/mK
Power through the oven wall dQ/dt = k * Area * temperature_difference / thickness
So total energy Q = k * Area * temperature_difference * time / thickness
Check the units, J = J/smK * m^2 * K * s / m

k = 0.045 W/mK
A = 1.8 m^2
deltaT = 107 K
time = 7*3600 = 25,200s
L = 0.023m
 
  • #7
mgb_phys said:
Conductivity has units W/mK
Power through the oven wall dQ/dt = k * Area * temperature_difference / thickness
So total energy Q = k * Area * temperature_difference * time / thickness
Check the units, J = J/smK * m^2 * K * s / m

k = 0.045 W/mK
A = 1.8 m^2
deltaT = 107 K
time = 7*3600 = 25,200s
L = 0.023m

Why is your k in W/mK? Shouldn't it be in J/(smC)?

BTW, did you get 9496017.3913 J as your answer?
 
  • #8
keemosabi said:
Why is your k in W/mK? Shouldn't it be in J/(smC)?
It's the same thing, W = J/s and it is conventional to quote temperature differences in kelvin rather than deg C. They are the same size so it doesn't change the value - it's just a detail thing.

Also note the number of figures in your answer.
You are only told the area to two significant figures so you cannot possibly give an answer with 12digits of accuracy. You should put 9.5 x106 or 9,500,000 J
 
  • #9
mgb_phys said:
It's the same thing, W = J/s and it is conventional to quote temperature differences in kelvin rather than deg C. They are the same size so it doesn't change the value - it's just a detail thing.

Also note the number of figures in your answer.
You are only told the area to two significant figures so you cannot possibly give an answer with 12digits of accuracy. You should put 9.5 x106 or 9,500,000 J

It worked! Thank you for the help.
 

1. How is energy usage measured in an oven?

Energy usage in an oven is measured in watts (W) or kilowatt-hours (kWh). Watts measure the rate at which energy is used, while kilowatt-hours measure the total amount of energy used over a period of time.

2. What factors affect the energy usage of an oven?

The energy usage of an oven can be affected by several factors, including the type of oven (electric or gas), the size and age of the oven, the temperature and cooking time, and the energy efficiency of the oven.

3. How much energy is typically used by an oven?

The amount of energy used by an oven can vary greatly depending on the factors mentioned above. However, on average, an electric oven typically uses around 2,000 watts (or 2 kWh) per hour, while a gas oven uses around 500 watts (or 0.5 kWh) per hour.

4. How much energy does operating an oven for 7 hours use?

To calculate the energy usage for operating an oven for 7 hours, you would need to know the wattage of the oven and the cost of electricity in your area. For example, if you have an electric oven that uses 2,000 watts per hour and your electricity rate is $0.15 per kWh, then operating the oven for 7 hours would use 14 kWh of energy, costing approximately $2.10.

5. How can I reduce the energy usage of my oven?

To reduce the energy usage of your oven, you can consider using smaller appliances such as a toaster oven or microwave for smaller meals, cooking multiple dishes at once to maximize oven space, using the oven's self-cleaning feature sparingly, and ensuring proper insulation and sealing of the oven door to prevent heat loss.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Back
Top