A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.10 m2 and whose thickness is 8 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 11 m2 and 0.15 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window

the percentage of heat lost by the window is 93.18%

Explanation:

Given the data in the question;

Area of glass = 0.10 m²

Thickness of glass = 8 mm = 0.008 m

Area of Styrofoam = 11 m²

Thickness of Styrofoam = 0.15 m

we know that;

Thermal conductivity of glass =  0.80 J/smC°

Thermal conductivity of Styrofoam   =  0.010 J/smC°

Now, temperature difference between outside and inside the walls and window is ΔT

So, In time t, heat lost due to conduction in the window will be;

= [ ×  × ΔTt] /

we substitute

= [ 0.80 × 0.10 × (ΔT)t] / 0.008

= [ 0.80 × 0.10 × (ΔT)t] / 0.008  

= 10(ΔT)t J

Also, the heat lost due to conduction in the wall be;

=  [ × × ΔTt] /

we substitute

=  [ 0.010 × 11 × ΔTt] / 0.15

=  0.7333(ΔT)t J

Now, Net heat lost in the wall and window is;

Q = +

Q = 10(ΔT)t J + 0.7333(ΔT)t J

Q = 10.7333(ΔT)t J

So, the percentage of heat lost by the windows will be;

% of heat lost = / Q

= 10(ΔT)t J / 10.7333(ΔT)t J

= 0.93167

= ( 0.93167 × 100 )%

= 93.18%

Therefore, the percentage of heat lost by the window is 93.18%

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.