Answer:
heat lost to the surrounding is 3429.30 J/s
Explanation:
given data
heat Q = 7200 J/s
surface temperature t1 = 400°C = 400+273 = 673 K
surroundings temperature t2 = 27°C = 27 + 273 = 300 K
surface area A = 0.45 m²
emissivity = 0.75
diameter = 44 cm
length = 25 cm
to find out
rate of heat lost to the surrounding
solution
we will apply here convection heat loss formula that is
heat Q = Q surface + Q surrounding ...........1
Q = σ ∈ A ( [tex]t1^{4} -t2^{4}[/tex] ) + Q surrounding
put here value
7200 = 5.67 ×[tex]10^{-8}[/tex] × 0.75 × 0.45 × ( [tex]673^{4} -300^{4}[/tex] ) + Q surrounding
solve it and we get Q surrounding
Q surrounding = 3429.30
so heat lost to the surrounding is 3429.30 J/s