In the figure (not to scale) a special wall has three metal studs with conductivity of 265 W/m·K. The wall otherwise has a conductivity of 0.88 W/m·K. What is the energy loss per second to the nearest watt if the temperature difference across the wall is 25 degrees Celsius?
To see the figure, please go to this website: http://wps.prenhall.com/esm_walker_physics_2/12/3138/803514.cw/index.html
Click on: "Practice Problems" on the left hand side (link)
Then scroll down to problem number 9
Thank you
I already gave this problem a shot
To see the figure, please go to this website: http://wps.prenhall.com/esm_walker_physics_2/12/3138/803514.cw/index.html
Click on: "Practice Problems" on the left hand side (link)
Then scroll down to problem number 9
Thank you
I already gave this problem a shot
-
Ambiguous, which way is the temperature difference applied? horizontally or vertically?
Although a 10 m thick wall is pretty big, it's not impossible. But I'll assume the 0.1 meter dimension is the direction of the heat flow. I also have to assume that the wall is uniform over it's 3 meter height. Both should have been stated explicitly.
Also problem has different numbers than the text in your question. I'll use the ones in the link.
look at it as 3 low value resistors (high conductivity) in parallel, and also in parallel is two resistors of low conductivity.
In parallel, conductances add. So first convert each section to conductance. We need the cross-sectional area, which is 0.6m x 3m = 1.8 m² and (10–0.6)m x 3m = 28.2 m²
conductances are
491 W/m•K x 1.8 m² = 884 Wm/K
1.53 W/m•K x 28.2 m² = 43 Wm/K
total is 927 Wm/K
927 Wm/K x 25K / 0.1m = 232000 watts (or 232000 J/s)
.
Although a 10 m thick wall is pretty big, it's not impossible. But I'll assume the 0.1 meter dimension is the direction of the heat flow. I also have to assume that the wall is uniform over it's 3 meter height. Both should have been stated explicitly.
Also problem has different numbers than the text in your question. I'll use the ones in the link.
look at it as 3 low value resistors (high conductivity) in parallel, and also in parallel is two resistors of low conductivity.
In parallel, conductances add. So first convert each section to conductance. We need the cross-sectional area, which is 0.6m x 3m = 1.8 m² and (10–0.6)m x 3m = 28.2 m²
conductances are
491 W/m•K x 1.8 m² = 884 Wm/K
1.53 W/m•K x 28.2 m² = 43 Wm/K
total is 927 Wm/K
927 Wm/K x 25K / 0.1m = 232000 watts (or 232000 J/s)
.