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If only conductive heat loss was significant for a house, then by what percentage would you lower the heat loss if the temperature was reduced inside from 70 F to 60 F when the outside temperature is 20 F?

Respuesta :

syed514
Let's try and solve this using the heat conductivity equation.Q=−kA(ΔT)/(Δx)

Where Q represents heat loss, k represents thermal conductivity, A represents area of the wall, and x is the thickness of the wall.
We'll assume everything except Q and delta T is for this case only, 1.
Case 1 : 70 F - 20 F = 50 F.Q=−1×1(50)/(1)
That's -50. Negative because energy is being transferred outside.Assuming the same trend..
Case 2 : 60F - 20F = 40F difference.I'm going to skip the calc. for this one, since it's pretty obvious. You're ending up with -40 as the answer here.
Now, let's express the answer of Case 2 where the temp difference is -40, as a percentage of Case 1. 
−40−50×100=80

Therefore, you're lowering the heat loss by 20%. 
okpalawalter8

We have that the heat loss is mathematically reduced by

Ph=20%.

Conductive heat loss

Question Parameters:

if the temperature was reduced inside from 70 F to 60 F when the outside temperature is 20 F

Generally the equation for the heat conductivity  is mathematically given as

.Q=−kA(ΔT)/(Δx)

Where

70 F - 20 F = 50 F.Q=−1×1(50)/(1)

-50. Negative because energy is being transferred out.

Hence

60F - 20F = 40F

-40. Negative because energy is being transferred out.

Finding the Percentage

 −40/−50×100=80 %

Giving

Ph=100-89=20%

Therefore,  heat loss is reduced by 20%.

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