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

We know the First Law of thermodynamics,
The first law of thermodynamics states that energy can be changed over starting with one structure then onto the next yet it can't be made nor devastated.
The numerical expression is,
$\Delta E = - q + w$
Where,
$\Delta E$ is the change in internal energy.
q is the amount of heat exchanged.
w is the work done by the system.

Complete step by step answer:
We all know the numerical expression of first law thermodynamics,
$\Delta E = q + w$
An adiabatic cycle is one in which no warmth is picked up or lost by the framework. The principal law of thermodynamics with $q = 0$ shows that all the change in interior energy is as work done. Therefore, from the first law thermodynamics,
$\Delta E = q + W$
On substituting the value of q we get,
$\Rightarrow \Delta E = 0 + W$
$\Delta E = W$
Therefore the change in internal energy is equal to the amount of work done on the system but it is not equivalent to zero.

So, the correct answer is Option A.

Note: One can determine the internal energy for one mole of carbon monoxide using the first law of thermodynamics.
Example:
Let us assume,
The total heat capacity of the calorimeter $C = 3kJ{\left( {^oC} \right)^{ - 1}}$
The initial temperature ${T_1} = {22.113^o}C$
The final temperature ${T_2} = {22.799^o}C$
The change in temperature is given by,
$\Delta T = {T_2} - {T_1}$
On substituting the known values we get,
$\Rightarrow \Delta T = {22.799^o}C - {22.113^o}C$
On simplifying we get,
$\Rightarrow \Delta T = {0.686^o}C$
The change in temperature is ${0.686^ \circ }C$
The amount of heat interchange can be estimated as,
$q = C\Delta T$
$q = 3kJ{\left( {^oC} \right)^{ - 1}} \times {0.686^o}C$
$q = 2.058kJ$
${\text{The heat released}} = {\text{The heat gained by the calorimeter}}$
The quantity of heat get by the calorimeter is $- 2.058kJ$
In bomb calorimeter the volume is fixed therefore the work done by the calorimeter is given by,
$w = P\Delta V$
$w = 0$
The work done by the calorimeter is zero.
The internal energy change can be calculated as,
$\Delta E = q + w$
$\Delta E = - 2.058kJ$
The internal energy change is $- 2.058kJ$
The internal energy for one mole of carbon monoxide can be calculated as,
$\Delta E = \dfrac{{ - 2.058kJ}}{{1.40g}} \times \dfrac{{38g}}{{1mol}} = - 41.16kJ/mol$
The internal energy for one mole of carbon monoxide is $- 41.16kJ/mol$.