Respuesta :
Answer:
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
Answer:
Internal energy ∆U=-26KJ
Explanation:
Given that:
Work done by the motor=+30KJ
Heat gained by the motor=+4KJ
In solving thermodynamical questions it is reasonable to use the sign convention this
Heat is positive if it is added to a system,but becomes negative if the system rejects heats.
Work is positive if the system does work,but becomes negative if work is done on the system.
Using the thermodynamics first law
∆U=Q-W
∆U= 4-30=-26KJ
