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

[keemosabi]

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?

 

[Delphi51]

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.

 

[keemosabi]

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?

 

[mgb_phys]

What do you mean by the 7 hours?

You can't multiply hours by J/(s·m·°C).

 

[keemosabi]

You can't multiply hours by J/(s·m·°C).

So convert it to seconds? I did that and still got the wrong answer.

 

[mgb_phys]

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

 

[keemosabi]

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?

 

[mgb_phys]

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

 

[keemosabi]

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.