Answer
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Hint First of all let us find out the change in internal energy. This change in internal energy is equal to the difference between the total heat energy and the work done by the system. The work done on the system is mentioned as negative and energy released is mentioned as negative. These all may help you to solve this question.
Complete Step-by-step answer
The change in internal energy is expressed as,
$\Delta U = {U_{final}} - {U_{initial}}$
Following the first law of thermodynamics, change in internal energy is equal to the difference of the total heat energy and the work done by the system. This can be expressed mathematically as,
$\Delta U = Q - W$
Therefore, we can equate both the equations together,
${U_{final}} - {U_{initial}} = Q - W$
As stated in the question, the heat energy has been released from the system.
According to the conventions, the heat energy released is specified as negative and vice-versa.
Therefore we can write that,
$Q = - Q$
And the work done is mentioned as work done on the system. As per the conventions, the work done on the system is mentioned as negative. Therefore we can write that,
$W = - W$
Hence we can substitute this in the equation,
That is,
${U_{final}} - {U_{initial}} = - Q + W$
The value of work done of the gas is given as,
$W = 8J$
The heat energy released is given as,
$Q = 20J$
The initial internal energy is given as,
${U_{initial}} = 30J$
Substituting these values in the equation will give,
${U_{final}} - 30 = - 20 + 8$
${U_{final}} = 18J$
So, the correct answer is (C) $18J$.
Note The first law of thermodynamics is defined as a version of the law of conservation of energy. When it is taken for the thermodynamic processes, we can distinguish this into two. One is the transfer of energy as heat and the other one is the transfer of energy in the form of work.
Complete Step-by-step answer
The change in internal energy is expressed as,
$\Delta U = {U_{final}} - {U_{initial}}$
Following the first law of thermodynamics, change in internal energy is equal to the difference of the total heat energy and the work done by the system. This can be expressed mathematically as,
$\Delta U = Q - W$
Therefore, we can equate both the equations together,
${U_{final}} - {U_{initial}} = Q - W$
As stated in the question, the heat energy has been released from the system.
According to the conventions, the heat energy released is specified as negative and vice-versa.
Therefore we can write that,
$Q = - Q$
And the work done is mentioned as work done on the system. As per the conventions, the work done on the system is mentioned as negative. Therefore we can write that,
$W = - W$
Hence we can substitute this in the equation,
That is,
${U_{final}} - {U_{initial}} = - Q + W$
The value of work done of the gas is given as,
$W = 8J$
The heat energy released is given as,
$Q = 20J$
The initial internal energy is given as,
${U_{initial}} = 30J$
Substituting these values in the equation will give,
${U_{final}} - 30 = - 20 + 8$
${U_{final}} = 18J$
So, the correct answer is (C) $18J$.
Note The first law of thermodynamics is defined as a version of the law of conservation of energy. When it is taken for the thermodynamic processes, we can distinguish this into two. One is the transfer of energy as heat and the other one is the transfer of energy in the form of work.
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