# 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 $\left(∆u\right)$ and the work done $\left(∆W\right)$ 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

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