Answer:
The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²
Explanation:
Here we have the heat Q given as follows;
Q = 15 × 1075 × (1100 -  ) = 10 × 1075 × (850 - 300) = 5912500 J
) = 10 × 1075 × (850 - 300) = 5912500 J
∴ 1100 -  = 1100/3
 = 1100/3
  = 733.33 K
  = 733.33 K

Where
 = Arithmetic mean temperature difference
 = Arithmetic mean temperature difference
 = Inlet temperature of the gas = 1100 K
 = Inlet temperature of the gas = 1100 K
 = Outlet temperature of the gas = 733.33 K
 = Outlet temperature of the gas = 733.33 K 
 =  Inlet temperature of the air = 300 K
 =  Inlet temperature of the air = 300 K
 = Outlet temperature of the air = 850 K
 = Outlet temperature of the air = 850 K
Hence, plugging in the values, we have;

Hence, from;
 , we have
, we have
5912500  = 90 × A × 341.67
 
Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².
 
        
             
        
        
        
Answer:
Temperature on the inside ofthe box 
Explanation:
The power of the light bulb is the rate of heat conduction of the bulb, 
The thickness of the wall, L = 1.2 cm = 0.012m
Length of the cube's side, x = 20cm = 0.2 m
The area of the cubical box, A = 6x²
A = 6 * 0.2² = 6 * 0.04
A = 0.24 m²
Temperature of the surrounding, 
Temperature of the inside of the box, 
Coefficient of thermal conductivity, k = 0.8 W/m-K
The formula for the rate of heat conduction is given by:
 
 
 
        
             
        
        
        
Answer:
I know this sounds quite deep but it is as deep as a grave 
Explanation:
It's reality 
 
        
                    
             
        
        
        
Answer:
can't understand the writting
 
        
                    
             
        
        
        
Answer:
c,d,e
Explanation:
rude, so you can keep safe and so you can really learn