Ahot engine generates excess heat at a rate of 7200 j per second. it has a surface temperature of 400°c, while its surroundings are at 27°c. the engine has a surface area of 0.45 m^ 2 and an emissivity of 0.75. the engine also loses heat via a steel conduit of diameter 44 cm and length 25 cm. assuming the temperature of the engine is stable, at what rate is heat lost to the surrounding air via convection?

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 ( ) + Q surrounding  

put here value

7200 = 5.67 × × 0.75  × 0.45 × ( ) + Q surrounding  

solve it and we get Q surrounding  

Q surrounding   = 3429.30

so heat lost to the surrounding is 3429.30 J/s

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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