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: 11
        A perfect carnot engine utilizes an ideal gas the source temperature is500K and sink temperature is 375K. If the engine takes 600k cal per cyclefrom the source, calculate(i) The efficiency of engine(ii) Work done per cycle(iii) Heat rejected to sink per cycle.
one year ago

Answers : (1)

Arun
20276 Points
							
Dear Vaibhav
 
T1 = 500 K
T2 = 375 K
Q1 = heat absorbed = 600 k cal
i) efficiency (n) = 1 -(T2/T1) = 125/500 = 0.25
 = 25%
ii) n = W/Q1
W = n* Q1 = 0.25 * 600 = 150 k cal
iii)- 
W = Q1 - Q2
Q2 = Q1 - W
Q2 = 600 - 150 = 450 k cal
 
Regards
Arun (askIITians forum expert)
one year 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