Question

When the heat transfer into a system is more than the work transfer out of the system, then:
A) The internal energy of the system remains constant
B) The internal energy of the system decreases
C) The internal energy of the system increases
D) None of the above

Answer 1

Hint: In order to answer this question, we have to understand and use the concept of First law of thermodynamics which states that the total energy of an isolated system is constant and energy can be transformed from one form to another.

Complete answer:
The first law of thermodynamics provides the definition of three important terms: work, heat and internal energy.
Work is the form of energy that is transferred by the application of the mechanical force resulting in a displacement.
Heat is the form of energy that is transferred by the virtue of the difference in the temperature between the bodies.
As per the first law of thermodynamics, heat and work are interchangeable. It means that heat added to a system can be converted to the work done by the system and the work done on a system can be converted to heat energy.
However, the heat energy and the work are transactional with respect to a system. The system has an innate energy known as the internal energy which is the total energy of the system which includes the potential energy, kinetic energy and the vibrational energy of the molecules of the system.
If heat is added to a system, a part of it is used to increase the internal energy of the system and the remaining heat is used in performing work.
Mathematically, if $\Delta Q$ is the heat added to the system and $\Delta W$ is the work done by the system and $\Delta U$ is the change in the internal energy of the system due to heat added, as per the First law of thermodynamics, we have –
$\Rightarrow \Delta Q = \Delta U + \Delta W$
Let us consider a case when heat added is more than the work done in the system –
$\Rightarrow \Delta Q > \Delta W$
In that case, the difference between the heat and work will be a positive number.
$\Rightarrow \Delta Q - \Delta W = + \Delta U$
Since their difference is equal to change in internal energy, we can say that the internal energy increases since it has a positive sign.

Hence, the correct option is Option C.

Note: This is only applicable for open and closed systems because heat and work can be easily exchanged in them. However, in adiabatic processes, the heat added to the system is zero. Thus,
When $\Delta Q = 0$ , we have –
$\Delta W = - \Delta U$
Thus, in absence of heat addition to the system, the work done is at the expense of loss of the internal energy in the case of an adiabatic system.