Question

At ${27^ \circ }C$ latent heat of fusion of a compound is 2930 J/mol. Entropy change is:
 (A) 9.77 J/mol K
(B) 10.77 J/mol K
(C) 9.07 J/mol K
(D) 0.977 J/mol K

Answer 1

Hint: The relation between the entropy change with the absolute temperature of the system and the heat can be given as
\[\Delta S = \dfrac{{{q_{rev}}}}{T}{\text{ }}\]

Complete step by step solution:
We are given the latent heat of fusion of a particular compound. Latent heat of the fusion is the heat we need to give to a compound to make the transition from solid to liquid state occur at constant pressure.
We are also provided the temperature at which the process takes place. Entropy of the system is the degree of randomness present in the system.
So, we can directly find the entropy change from its formula. The formula can be given as
\[\Delta S = \dfrac{{{q_{rev}}}}{T}{\text{ }}....{\text{(1)}}\]
Here, ${q_{rev}}$ is the heat and we can put the value of latent heat of the compound here. T is the absolute temperature.
We are given that T = ${27^ \circ }C$. Now, we require the absolute temperature which is in the Kelvin unit.
We know that $^ \circ C$ =273+K
Thus, ${27^ \circ }C$= 27+273K = 300 K
Also we are given that ${q_{rev}}$ =2930 J/mol
so putting it in the equation (1) will give
\[\Delta S = \dfrac{{2930}}{{300}} = 9.766Jmo{l^{ - 1}}{K^{ - 1}}\]
Thus, we obtained that the entropy change for the given process is 9.766 J/mol K.

Therefore, the correct answer is (A).

Note: Remember that we always need to put the temperature in the Kelvin unit in this equation of entropy change. Putting the temperature in Celsius units will lead to error. This latent heat of fusion of the compound can also be regarded as the enthalpy of fusion. Do not forget to write the required units behind the value of the property.