Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        A spherical ball of surface area 20 cm2 absorbs any radiation that falls on it. It is suspended in a closed box maintained at 57°C. (a) Find the amount of radiation falling on the ball per second. (b) Find the net rate of heat flow to or from the ball at an instant when its temperature is 200°C. Stefan constant = 6.0 × 10-8 W/m2-K4.`
5 years ago

```							Sol. (a) A = 20 cm^2 = 20× 10^–4 m^2, T = 57°C = 330 K
E = A σT^4 = 20 × 10^–4 × 6 × 10^–8 × (330)^4 × 10^4 = 1.42 J
(b) E/t = Aσe(T base 1^4 – T base 2^4),  A = 20 cm^2 = 20 * 10^-4 m^2
σ = 6 * 10^-8 T base 1 = 473 K, T base 2 = 330 K
= 20 × 10^–4 × 6 × 10^–8 × 1[(473)^4 – (330)^4]
= 20 × 6 × [5.005 × 10^10 – 1.185 × 10^10]
= 20 × 6 × 3.82 × 10^–2 = 4.58 w from the ball.

```
5 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Thermal Physics

View all Questions »  ### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions