# Energy Question-(don't know which equation to use?)

1. Oct 4, 2007

### quickclick330

Energy Question--(don't know which equation to use?)

I don't know how to start the problem, any help or hints would be greatly appreciated.

I'm gussing you are supposed to use Ethermal = mCdT but not sure how to apply it to this question. Thanks again!!:-)

During 4 hours one winter afternoon, when the outside temperature was 5° C, a house heated by electricity was kept at 23° C with the expenditure of 49 kwh (kilowatt·hours) of electric energy.

(a) What was the average energy leakage in joules per second (watts) through the walls of the house to the environment (the outside air and ground)?

(b)The rate at which energy is transferred between two systems due to a temperature difference is often proportional to their temperature difference. Assuming this to hold in this case, if the house temperature had been kept at 26° C (77° F), how many kwh of electricity would have been consumed?

2. Oct 4, 2007

### learningphysics

Assuming all the electricity is converted to heat... how much energy leaves the house... remember the temperature is remaining constant at 23 degrees...

You have the energy... divide by time to get rate of energy leakage.

3. Oct 4, 2007

### quickclick330

i calculated energy to be 49kw/h = 49000 w/h = 49000 J / (4 hours X 60) = 204.1
is that right?

4. Oct 4, 2007

### learningphysics

No. you're given 49kwh. kwh is a unit of energy not power... kw*h not kw/h... you can convert kwh to joules if you want. But you don't need to...

Divide energy by time... ie divide 49kwh by the time...

5. Oct 4, 2007

### quickclick330

Solved! Thanks for the help :-)

6. Oct 4, 2007

### solrac214

I have the exact same problem but I don't understand the energy principal just yet (hasn't clicked) I can't get the answer for part a

7. Oct 4, 2007

### solrac214

I ran out of guesses/failed :( but I still need to learn how to do that for the test so anymore help would be appreciated

8. Oct 5, 2007

### learningphysics

The temperature is kept constant. The idea is that all of the 49kwh is converted to heat... but if that is the case all that heat must leave the house... reason being that the temperature in the house is constant... so heat entering house - heat leaving the house = 0.

So 49kwh is lost over 4 hours... so the rate at which energy is lost if 49kwh/4h = 12.25kw.

9. Oct 5, 2007

### solrac214

tried that and it wasn't the right answer =(

10. Oct 5, 2007

### learningphysics

really? did you need to enter the answer in W or kW?

oh... the question asks for watts... so 12250W...

11. Oct 5, 2007

### solrac214

my friend had the numbers 4 hrs 4 degrees outside 22 degrees inside and 51 kwh and his answer was 1417.17 watts if that helps but someone else solved it for him

12. Oct 5, 2007

### learningphysics

Hmmm... I'm curious how this problem should be solved then...