A cubical block of mass 1.0 kg and edge 5.0 cm is heated to 227°C. It is kept in an evacuated chamber maintained at 27°C. Assuming that the block emits radiation like a blackbody, find the rate at which the temperature of the block will decrease. Specific heat capacity of the material of the block is 400 J Kg-1 K-1.
It is given that a cube behaves like a black body.
∴ Emissivity, e = 1
Stefan's constant, σ = 6 × 10−8 W/(m2 K4)
Surface area of the cube, A = 6 × 25 × 10−4
Mass of the cube, m = 1 kg
Specific heat capacity of the material of the cube, s = 400 J/(kg-K)
Temperature of the cube, T1 = 227 + 273 = 500 K
Temperature of the surrounding, T0 = 27 + 273 = 300 K
Rate of flow of heat is given by
`(DeltaQ)/(Delta)`=eAσ ( T4 -T0 4)
⇒ ms . `(DeltaT)/(Deltat)` = 1× 6 × 10 × 25 × 10-4 (5004 -3004)
`rArr (DeltaT)/(Deltat) = (36xx25xx(500^4 - 300^4)xx10^-12)/400`
⇒ `(DeltaT)/(Deltat)= 0.12^circ C/s`
Video Tutorials For All Subjects
- Qualitative Ideas of Blackbody Radiation