Concepts Of Physics MCQ Edition [Volume 2]PhysicsHeat Transfer
Steam at 120^ C is continuously passed through a rubber tube that is 50 cm long with inner and outer radii of 1.0 cm and 1.2 cm , respectively. The room temperature is 30^ C . Determine the rate of heat flow through the walls of the tube. The thermal conductivity of rubber is 0.15 J s ⁻¹ m ⁻¹ ^ C⁻¹ .
Options
- A233 J s ⁻¹
- B388 J s ⁻¹
- C212 J s ⁻¹
- D254 J s ⁻¹
Correct answer
A. 233 J s ⁻¹
Step-by-step solution
The rate of heat flow through a cylindrical wall is given by: H = 2 K L (T₁ - T₂) (r₂/r₁) Substituting the given values: K = 0.15 J s ⁻¹ m ⁻¹ ^ C ⁻¹ L = 0.5 m T₁ = 120^ C , T₂ = 30^ C r₁ = 1.0 cm , r₂ = 1.2 cm H = 2 (0.15) (0.5) (120 - 30) (1.2/1.0) H = 0.15 (90) (1.2) Using (1.2) 0.1823 and 3.1416 : H = 13.5 3.1416 0.1823 H 232.6 J s ⁻¹ Rounding to the nearest integer gives 233 J s ⁻¹ .