NEET Physics Thermodynamics Questions Solved

NEET - 2009

The internal energy change in a system that has absorbed 2 kcal of heat and done 500 J of work is 

(a) 8900 J                                      (b) 6400 J

(c) 5400 J                                      (d) 7900 J

Heat given to a system (Q) is equal to the sum of increase in the internal energy u and the work done W by the system against the surrounding and 1 cal=4.2J.

According to first law of thermodynamics

            U=Q-W

                 = 2×4.2×1000-500

                = 8400-500

                = 7900 J

Difficulty Level:

  • 14%
  • 18%
  • 17%
  • 53%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 25, 2019)