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
Menu
Grade: upto college level
        A body cools down from 50°C to 45°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surrounding. 
5 years ago

Answers : (1)

Deepak Patra
askIITians Faculty
471 Points
							Sol. 50°C 45°C 40°C
Let the surrounding temperature be ‘T’°C
Avg. t = 50 + 45/2 = 47.5
Avg. temp. diff. from surrounding
T = 47.5 – T
Rate of fall of temp = 50 – 45/5 = 1 °C/mm
From Network’s Law
1°C/mm = bA * t
⇒ bA = 1/t = 1/47.5 – T …(1)
In second case,
Avg, temp = 40 + 45/2 = 42.5
Avg. temp. diff. from surrounding
t’ = 42.5 – t
rate of fall of temp = 45 – 40/8 = 5/8 °C/mm
From Network’s Law
⇒ 5/B = 1/(47.5 - T) * (42.5 - T)
By C & D [Componendo & Dividendo method]
We find, T = 34.1°C

						
5 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


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


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details