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 = ![\frac{7800008400 \ Watt-Hour}{3650 \ Hours} = 2136988.6 \ W](https://tex.z-dn.net/?f=%5Cfrac%7B7800008400%20%5C%20Watt-Hour%7D%7B3650%20%5C%20Hours%7D%20%20%3D%202136988.6%20%5C%20W)
Rate of heat transfer, ![P = \frac{K*A *\delta T}{L}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BK%2AA%20%2A%5Cdelta%20T%7D%7BL%7D)
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)
![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](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BKA%28T_2-T_1%29%7D%7BL%7D%20%5C%5C%5C%5CL%20%3D%20%20%5Cfrac%7BKA%28T_2-T_1%29%7D%7BP%7D%5C%5C%5C%5CL%20%3D%20%20%5Cfrac%7B385%284.5%2A3.84%29%2850-30%29%7D%7B2136988.6%7D%5C%5C%5C%5CL%20%3D%200.0623%20%5C%20m)
L = 6.23 cm
Therefore, the thickness of the material is 6.23 cm