Question

What will be the change in internal energy when ${ 12kJ }$ of work is done on the system and ${ 2kJ }$ of heat is given by the system?
A. ${ +10kJ }$
B. ${ -10kJ }$
C. ${ +5kJ }$
D. ${ -5kJ }$

Answer 1

Hint: Internal energy of any system is the total of all the kinds of energy present in the system. It does not depend on the path but only on the state of the system and hence it is a state function.
According to the first law of thermodynamics, ${ \Delta U=q+w }$

Complete step by step solution:
It is given that,
The heat evolved = ${ -2kJ }$ [Negative sign comes as heat is given by the system]
Work done on the system = ${ 12kJ }$
As we know, ${ \Delta U=q+w }$.............(1)
where, ${ \Delta U }$ = change in internal energy
q = heat evolved
w = work done on the system
Now, put the values in equation (1), we get
${ \Delta U } = { (-2)+12 }$
${ \Delta U } = { +10kJ }$
Hence, the change in internal energy is ${ +10kJ }$.

The correct option is A.

Additional Information:
1. In thermodynamics, a state function, function of the state, state quantity, or state variable is known to be a property of a system that depends only on the present state of the system, or state variable is known to be a property of a system that depends only on the present state of the system, not on the way in which the system acquired that state (independent of the path).
2. A state function is a function that describes the equilibrium state of a system.
3. Internal energy is a state function.

Note: The possibility to make a mistake is that the negative sign comes when the heat evolves and the positive sign when the heat is given to the system. Don’t confuse between the two signs.