Answer:
Explanation:
The rate of conductive heat transfer in watts is:
q = (k/s) A ΔT
where k is the heat conductivity, s is the thickness, A is the area, and ΔT is the temperature difference.
a)
Given k = 52.4 W/m/K, s = 0.014 m, and ΔT = 350-140 = 210 K, we can find q/A:
q/A = (52.4 / 0.014) (210)
q/A = 786,000 W/m²
b)
Given that A = 0.42 m², we can find q:
q = (0.42 m²) (786,000 W/m²)
q = 330,120 W
A watt is a Joule per second. Convert to Joules per hour:
q = 330,120 J/s * 3600 s/hr
q = 1.19×10⁹ J/hr
c)
If we change k to 1.8 W/m/K:
q = (k/s) A ΔT
q = (1.8 / 0.014) (0.42) (210)
q = 11,340 J/s
q = 4.08×10⁷ J/hr
d)
If k is 52.4 W/m/K and s is 0.024 m:
q = (k/s) A ΔT
q = (52.4 / 0.024) (0.42) (210)
q = 192,570 J/s
q = 6.93×10⁸ J/hr
