NTA Abhyas JEE Main2020PhysicsThermal Properties of MatterPractice
A black body of mass 34 . 38 g and surface area 19 . 2 cm 2 is at an initial temperature of 400 K . It is allowed to cool inside an evacuated enclosure kept at constant temperature 300 K . The rate of cooling is 0 . 04 ° C s - 1 . The specific heat of the body in J k g - 1 K - 1 is (Stefan's constant σ = 5.73 × 10 - 8 W m - 2 K - 4 )
Options
- A2800
- B2100
- C1400
- D1200
Correct answer
C. 1400
Step-by-step solution
According to Newtons's law of cooling d θ d t = σ A ( T 4 - T 0 4 ) m s ∴ Specific heat s = σ A ( T 4 - T 0 4 ) m d θ d t Substituting the values ∴ s = 5.73 × 10 - 8 19.2 × 10 - 4 [ 4 4 - 3 4 ] × 10 8 34.38 × 10 - 3 ( 4 × 10 - 2 ) = 1400