
If heat given to a system is $6Kcal$ and work done is $6KJ$. Then change in internal energy is: -
A. $19.1KJ$
B. $12.5KJ$
C. $25KJ$
D. $Zero$
Answer
473.4k+ views
Hint: We will use the first law of thermodynamics which states that heat energy cannot be created nor be destroyed. It can only be converted from one form to the other. Also, we will convert the unit $Kcal$ into $KJ$ since the given options of the answers are in $KJ$ . The formula which we will use here is $\Delta Q = \Delta U + \Delta W$.
Complete Step by step answer:
According to the first law of thermodynamics “The First Law says that the internal energy of a system has to be equal to the work that is being done on the system, plus or minus the heat that flows in or out of the system and any other work that is done on the system”
So, the formula used will be: -
$\Delta Q = \Delta U + \Delta W$
Or
$\Delta U = \Delta Q - \Delta W$
Where, $\Delta U$is the change in internal energy, $\Delta Q$ is the change in heat energy and $\Delta W$ is the work done.
First, we will convert $Kcal$ into $KJ$
$\Delta Q = 6Kcal$
Since $1Kcal = 4.2kJ$so, $6Kcal$ will be equal to: -
$\Delta Q = 6 \times 4.2KJ = 25.2KJ$
Now we will put the values in the formula
$
\Delta Q = 25.2KJ \\
\Rightarrow \Delta W = 6KJ \\
\Rightarrow \Delta U = \Delta Q - \Delta W \\
\Rightarrow \Delta U = 25.2KJ - 6KJ \\
\Rightarrow \Delta U = 19.2KJ \\
$
The system will experience a $19.2KJ$ of increase in internal energy.
So, the correct option will be A. i.e $19.1KJ$ as it is the closest to the answer calculated.
Note: It should be noted that the internal energy of a system would decrease if the system gives off heat or does work. Therefore, internal energy of a system increases when the heat increases (this would be done by adding heat into a system). The internal energy would also increase if work were done onto a system. Any work or heat that goes into or out of a system changes the internal energy.
Complete Step by step answer:
According to the first law of thermodynamics “The First Law says that the internal energy of a system has to be equal to the work that is being done on the system, plus or minus the heat that flows in or out of the system and any other work that is done on the system”
So, the formula used will be: -
$\Delta Q = \Delta U + \Delta W$
Or
$\Delta U = \Delta Q - \Delta W$
Where, $\Delta U$is the change in internal energy, $\Delta Q$ is the change in heat energy and $\Delta W$ is the work done.
First, we will convert $Kcal$ into $KJ$
$\Delta Q = 6Kcal$
Since $1Kcal = 4.2kJ$so, $6Kcal$ will be equal to: -
$\Delta Q = 6 \times 4.2KJ = 25.2KJ$
Now we will put the values in the formula
$
\Delta Q = 25.2KJ \\
\Rightarrow \Delta W = 6KJ \\
\Rightarrow \Delta U = \Delta Q - \Delta W \\
\Rightarrow \Delta U = 25.2KJ - 6KJ \\
\Rightarrow \Delta U = 19.2KJ \\
$
The system will experience a $19.2KJ$ of increase in internal energy.
So, the correct option will be A. i.e $19.1KJ$ as it is the closest to the answer calculated.
Note: It should be noted that the internal energy of a system would decrease if the system gives off heat or does work. Therefore, internal energy of a system increases when the heat increases (this would be done by adding heat into a system). The internal energy would also increase if work were done onto a system. Any work or heat that goes into or out of a system changes the internal energy.
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