Question

A sample of 0.1g of water at $100^\circ C$and normal pressure $\left( {1.013 \times {{10}^5}N{m^{ - 2}}} \right)$requires 54 cal of heat energy of convert to steam at $100^\circ C$. If the volume of the steam produced is 167.1 cc, the change in the internal energy of the sample, is?
$\left( A \right)42.2J \\\ \left( B \right)104.3J \\\ \left( C \right)84.5J \\\ \left( D \right)208.7J \\\$

Answer 1

Hint: We will solve this question by first understanding it properly. We will note down all the given information in the question. Then we will state the First Law Of Thermodynamics and by using its mathematical formula, i.e., $\Delta Q = \Delta U + \Delta W$, we will get the required answer.

Complete step-by-step answer:

Formula used - $\Delta Q = \Delta U + \Delta W$

Here in the question, it is given that, a sample of 0.1g of water at $100^\circ C$ and normal pressure $\left( {1.013 \times {{10}^5}N{m^{ - 2}}} \right)$ requires 54 cal of heat energy of convert to steam at $100^\circ C$ and the volume of the steam produced is 167.1 cc.
Now, we have to find out the change in the internal energy of this sample.

We will state the First Law of Thermodynamics.
As we know that according to the first law of thermodynamics, “if some significant amount of heat is supplied in a system which can do some work, then the amount of heat absorbed in the system will be equal to the sum of the increase in its internal energy and the external work done by the system on the surrounding.”

Let,
$\Delta Q$ = Heat supplied to the system by the surroundings
$\Delta W$= Work done by the system on the surroundings
$\Delta U$ = Change in the internal energy of the system
Then according to the First Law of Thermodynamics,

$\Delta Q = \Delta U + \Delta W$

Here,

Heat supplied$\left( {\Delta Q} \right)$ = 54 cal = 54 $\times$4.2$J$= 226.8$J$
Work done$\left( {\Delta W} \right)$

We know that,
$\Delta W = P\Delta V \\\ \therefore \Delta W = 1.01 \times {10^5} \times 167.1 \times {10^6}J \\\ \Delta W = 168.8 \times {10^{ - 1}}J \\\ \Delta W = 16.88J \\\$

The change in internal energy of the sample $\left( {\Delta U} \right)$= ?

Now, by using First Law of Thermodynamics,

$\Delta Q = \Delta U + \Delta W$
$\Rightarrow \Delta U = \Delta Q - \Delta W \\\ \Rightarrow \Delta U = 226.8 - 16.88 \simeq 208.7J \\\ \therefore \Delta U = 208.7J \\\$

Therefore, the change in the internal energy of the sample is $208.7J$.

Hence, option D is the right answer.

Note – The internal energy of a system is the sum of molecular kinetic and potential energies in the frame of reference relative to which the centre of mass of the system is at rest. Heat is the mode of energy transfer due to temperature difference between the system and the surroundings. Work is defined as the mode of energy transfer brought about by means that do not involve temperature distinction.