Question

3000 J of heat is added to a system and 2500 J of work is done by the system. What is the change in internal energy of the system?​

Answer 1

Given:

3000 J of heat is added to a system and 2500 J of work is done by the system.

To find:

Internal energy change of system ?

Calculation:

Applying 1st Law of Thermodynamics:

$\Delta Q = \Delta U + W$

• Since heat was added to system, it sign will be positive.

• Also, since work was done by system, its sign will be positive.

$\implies 3000= \Delta U + 2500$

$\implies \Delta U = 3000 - 2500$

$\implies \Delta U = 500 \: joule$

So, internal energy change is 500 Joule.

Answer 2

Answer:

500J

Explanation:

GIVEN- 3000 J of heat is added to a system and 2500 J of work is done by the system.

TO FIND- the change in internal energy of the system

SOLUTION - according to first law of thermodynamics-

3000= ∆U+ 2500

∆U= 3000-2500

∆U= 500J

first law of thermodynamics- The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic processes, distinguishing three kinds of transfer of energy, as heat, as thermodynamic work, and as energy associated with matter transfer, and relating them to a function of a body's state, called internal energy

FINAl ANSWER- 500J

#SPJ2.