Answer:
The thickness of the material is 6.23 cm
Explanation:
Given;
quantity of heat, Q = 6706.8 *10⁶ kcal
duration of the heat transfer, t = 5 months
thermal conductivity of copper, k = 385 W/mk
outside temperature of the heater, T₁ = 30° C
inside temperature of the heater, T₂ = 50° C
dimension of the rectangular heater = 450 cm by 384 cm
1 kcal = 1.163000 Watt-hour
6706.8 *10⁶ kcal = 7800008400 watt-hour
I month = 730 hours
5 months = 3650 hours
Rate of heat transfer, P = [tex]\frac{7800008400 \ Watt-Hour}{3650 \ Hours} = 2136988.6 \ W[/tex]
Rate of heat transfer, [tex]P = \frac{K*A *\delta T}{L}[/tex]
where;
P is the rate of heat transfer (W)
k si the thermal conductivity (W/mk)
ΔT is change in temperature (K)
A is area of the heater (m²)
L is thickness of the heater (m)
[tex]P = \frac{KA(T_2-T_1)}{L} \\\\L = \frac{KA(T_2-T_1)}{P}\\\\L = \frac{385(4.5*3.84)(50-30)}{2136988.6}\\\\L = 0.0623 \ m[/tex]
L = 6.23 cm
Therefore, the thickness of the material is 6.23 cm
