Question

2000 calories of heat are given to a thermodynamis
system and the system does 3350 joule of external
work. In this process the internal energy of the system
is increased by 5030 joule. Calculate the value of the
conversion factor J.​

Answer 1

Answer:

The first law of thermodynamics defines the internal energy (E)

as equal to the difference of heat transfer (Q)

into the system (W)done by the system.....

E - E=Q-W

² ¹

In our case

∆E=E- E=5030J

². ¹

W=3350J

Q=∆E+W=5030+3350=8380J

given

Q=2000cal

Conversation factor J

K=8380/2000=4.19

so the answer is 1cal=4.19J #BAL#ANSWERWITHQUALITY

You are good brainly user....