# Thermal Conduction

1. Jan 12, 2009

### keemosabi

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Q = (k A deltaT time)/L

3. 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?

2. Jan 12, 2009

### 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.

3. Jan 12, 2009

### keemosabi

The units of the k is correct. I double checked it. What do you mean by the 7 hours?

4. Jan 12, 2009

### mgb_phys

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

5. Jan 12, 2009

### keemosabi

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

6. Jan 12, 2009

### 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

7. Jan 12, 2009

### keemosabi

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

8. Jan 12, 2009

### mgb_phys

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.