Amotor does 30 kj of work and gains 4 kj as heatfrom the surroundings. what is the change in the internal energy of the motor?

ΔU= *-26 KJ

Explanation:

Given that

Work done by motor W= 30 KJ

Heat gains by motor Q= 4 KJ

Sign convention:

 If heat is added to the system then it is taken as positive and if heat is rejected from the system then it is taken as negative.

If work done by the system then it is taken as positive and if work is done on the system then it is taken as negative.

From first law of thermodynamics

Q = W + ΔU

ΔU=Change in internal energy

Q=Heat transfer

W=Work

Now by putting the values

4 = 30 + ΔU

ΔU= -26 KJ

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.