Question

In a closed insulated container, a liquid is stirred with a paddle to increase the temperature, which of the following is true?
A. $\Delta E = \Delta W \ne 0,q = 0$
B. $\Delta E = \Delta W = 0,q \ne 0$
C. $\Delta E = 0,W = q \ne 0$
D. $W = 0,\Delta E = q \ne 0$

Answer 1

As the given question is conceptual we will use the basic concepts of thermodynamics. One of the hints is given in the question that is the system in a closed container. So we will understand the process and also observe the effect of temperature on the thermodynamic quantities. Based on the given situation we will observe the change in the thermodynamic quantities.

Formula Used: The first law of thermodynamics, $\Delta E = q + W$
Where $\Delta E$ represents the internal energy of the system, $q$ represents the heat of the system, $W$ represents the work done by/on the system.

Complete step by step answer:
First, we will understand the first law of thermodynamics. It states that the internal energy $\left( {\Delta E} \right)$ of the system is equal to the sum of heat absorbed/released of the system and the work done by/on the system. Now we will consider the conditions given in the question and relate it with the basic thermodynamic quantities.
It is given in the question, that the system is in a closed container. So we can conclude that there will be no heat transfer. Therefore the heat is $q = 0$. Now we have given that liquid is stirred with a paddle to increase the temperature which increases the internal energy $\Delta E$ of the system. As $\Delta E = nC\Delta T$ which implies that $\Delta E \ne 0$. Now work is also done on the system when stirred with a paddle which implies that $W \ne 0$. Now using the first law of thermodynamics, we get $\Delta E = q + W = 0 + W,\Delta E = W$. So now from our observation, we can conclude that $\Delta E = \Delta W \ne 0,q = 0$.

Therefore, the correct option is (A) $\Delta E = \Delta W \ne 0,q = 0$.

Note: If we express the first law of thermodynamics with the expression $q = \Delta E + W$. So, according to the conventions work done on the system is considered as negative. Therefore, $q = \Delta E - W$. Now in a closed container $q = 0$ which implies that $\Delta E = W$. So we can conclude from here that $\Delta E = \Delta W \ne 0,q = 0$