Heat transfer, energy needed to heat house

[Leaffy]

Hi everyone!
First time here, but I'm having trouble with a problem. Here goes:

On a given day, heat losses through the 20cm-thick walls of a warehouse are 0.02 kW/m2 when the outside wall temperature is 5C and the inside wall temperature is 25C. The owner would like to reduce her heating bill by adding urethane insulation to the outside wall. k:urethane = 0.026 W/m.K

a. If the walls have an area of 200 m2, what is the total energy (in kJ and kWh) needed to heat the house that day (a day has 24 hours) without the insulation (ignore windows and doors)
b. What is the thermal conductivity of the wall?
c. Determine the thickness of the insulation needed to reduce the heating bill that day by a factor of 2.


So the equation I'm looking at is Q=kA(ΔT/Δx)
I think I can use 0.02 kW/m2 multiplied by area to get a Wattage for power, but I don't think it's right.

Once I have power, I can do the rest, but this has been driving me nuts all day.

Your help is much appreciated.

 

[Redbelly98]

Welcome to Physics Forums!

So the equation I'm looking at is Q=kA(ΔT/Δx)
I think I can use 0.02 kW/m2 multiplied by area to get a Wattage for power, but I don't think it's right.

Actually that looks like the right approach to me.

 

[Leaffy]

Thank you, but what do I do about time? Or is that extraneous information?

 

[Redbelly98]

I think I can use 0.02 kW/m2 multiplied by area to get a Wattage for power...

Yes, you'll get the power, but question (a) is asking for energy. So you need to know the time to get the energy from the power that you calculate.

 

[asteriatic]

Hi there! I can help you with this problem. Let's break it down step by step.

a. To calculate the total energy needed to heat the house that day, we need to first calculate the heat loss through the walls. We can use the equation you mentioned, Q=kA(ΔT/Δx), where Q is the heat loss, k is the thermal conductivity, A is the area, ΔT is the temperature difference, and Δx is the wall thickness.

So, for the given values, we have Q = (0.02 kW/m2)(200 m2)(25C-5C)/0.2m = 2000 W = 2 kW.

To convert this to kJ, we multiply by the time in seconds (24 hours = 86,400 seconds) and we get 2 kW x 86,400 s = 172,800 kJ.

To convert this to kWh, we divide by 3,600 seconds (1 hour) and we get 172,800 kJ / 3,600 s = 48 kWh.

Therefore, the total energy needed to heat the house that day without insulation is 172,800 kJ or 48 kWh.

b. To find the thermal conductivity of the wall, we rearrange the equation to solve for k: k = Q/(AΔT/Δx). Plugging in the given values, we get k = 2 kW / (200 m2)(25C-5C)/0.2m = 0.02 W/m.K. Therefore, the thermal conductivity of the wall is 0.02 W/m.K.

c. To reduce the heating bill by a factor of 2, we need to reduce the heat loss by half. Using the same equation, we can solve for the new thickness of insulation needed (Δx') to achieve this: Δx' = Δx(Q/kA)/2. Plugging in the values, we get Δx' = (0.2m)(2 kW)/(0.026 W/m.K)(200 m2)/2 = 15.4 cm. Therefore, the thickness of the insulation needed to reduce the heating bill by a factor of 2 is 15.4 cm.

I hope this helps and clears up any confusion you had. Good luck with your problem!