Answer:
Q=1170J/s = 279.516cal/s
Step-by-step explanation:
q=-k.(T₀-T₁)/(x₀- x₁)
q= heat flow density
T= temperature
x= thickness
q= -(1.04 W/m. ºC) . (50ºC-500ºC)/(0.1m-0m)
q=4680W/m²=4,68KW/m²
1KW=1000J/s
∴ q= 4680 J/s.m²
q= Q/A
Q=heat fluw
A= surface= 6 . (0.5m . 0.5m)= 0.25m²
Q= (4680J/s.m²) . (0.25m²)
Q=1170J/s = 279.516cal/s
1J=0.2389cal