Question

The increase in the internal energy of 1 kg of water at $100^\circ C$ when it is converted into steam at the same temperature and at 1atm (100 kPa) will be [The density of water and steam are 1000 $\dfrac{{kg}}{{{m^3}}}$ and 0.6 $\dfrac{{kg}}{{{m^3}}}$ respectively. The latent heat of evaporation of water is $2.25 \times {10^6}$ J/kg]:
A. \[2.08 \times {10^6}\] J
B. $4 \times {10^7}$ J
C. \[3.27 \times {10^8}\] J
D. \[5 \times {10^5}\] J

Answer 1

Hint: The first law of thermodynamics states that the can be converted from one form to the other with the interaction of heat, work and internal energy, but under any circumstances, it can’t be created or destroyed.
 \[\Delta U = q + w\]

Complete step by step answer:
Latent heat of evaporation, i.e. q = $2.25 \times {10^6}$ J/Kg
Given,
Pressure, P = 100 kPa
P = $100 \times {10^3}$ P
So, by putting the values in formula
 \[W = - PV\]
 $W = - P\Delta V = - P({V_2} - {V_2})$
Now, Density $ = \dfrac{{Mass}}{{Density}}$| Or, Volume $ = \dfrac{{Mass}}{{Density}}$
Volume of steam, \[{V_2} = \dfrac{{Mass}}{{Density\,of\,steam}} = \dfrac{1}{{0.6}}\]
Similarly, Volume of Water, \[{V_1} = \dfrac{{Mass}}{{Density\,of\,steam}} = \dfrac{1}{{1000}}\]
Now putting the values in the formula
 $W = - P\Delta V = - P({V_2} - {V_1}) = - 100 \times {10^3}[\dfrac{1}{{0.6}} - \dfrac{1}{{1000}}$
 $W = - 1.67 \times {10^2}$

From the first law of thermodynamics,
 \[\Delta U = q + w\]
Where, \[\Delta U\] = Internal energy
q = Latent heat
W = Work
So, \[\Delta U = 2.25 \times {10^6} + ( - 1.67 \times {10^5}\]
 \[\Delta U = 2.25 \times {10^6} - 0.167 \times {10^5} = 2.08 \times {10^6}\]

Therefore, the correct answer is option (A).

Note: Latent heat of evaporation is q and change in internal energy is \[\Delta U\] and W is work and also work can be derived from the pressure and volume