What is the change in entropy when \(2.5\) mole of water is heated from \(27^{\circ} \mathrm{C}\) to \(87^{\circ} \mathrm{C} ?\) Assume that the heat capacity is constant. \(\left(C_{p, m}\left(\mathrm{H}_{2} \mathrm{O}\right)=4.2 \mathrm{~J} / \mathrm{g}-\mathrm{K} \ln (1.2)=0.18\right)\) (a) \(16.6 \mathrm{~J} / \mathrm{K}\) (b) \(9 \mathrm{~J} / \mathrm{K}\) (c) \(34.02 \mathrm{~J} / \mathrm{K}\) (d) \(1.89 \mathrm{~J} / \mathrm{K}\)

Thermodynamics is a branch of physics that deals with heat, work, and temperature, and their relation to energy, radiation, and physical properties of matter. It has several fundamental laws, such as the first law which introduces the concept of internal energy and equates changes in internal energy to heat added to a system minus work done by the system. The second law introduces the concept of entropy, a measure of the disorder or randomness in a system.

When dealing with thermodynamics problems, we often have to calculate work done by or on the system, changes in internal energy, or the heat exchanged. The change in entropy is particularly important because it provides a quantitative measure of the irreversibility in a given process. Understanding how to manipulate equations related to heat, temperature, and entropy is crucial for solving thermodynamics problems in a wide range of physical contexts.

Physical chemistry is the branch of chemistry that studies the physical properties of molecules, the forces that act upon them, and the interactions that lead to chemical reactions. It combines principles of physics and chemistry to understand the physical structure of molecules, the forces that act on them, and how they combine and react. Studies in physical chemistry involve understanding concepts like thermodynamics, reaction kinetics, quantum chemistry, and statistical mechanics.

In the context of entropy change calculations, physical chemistry provides an in-depth understanding of the molecular basis for why entropy changes occur. It delves into the atomic level, explaining that when substances are heated and their temperatures increase, the molecules gain more kinetic energy and move more vigorously, thus increasing the system's disorder or entropy.

The Joint Entrance Examination (JEE) is a highly competitive exam for students aiming to secure a place at prestigious Indian institutes for engineering and technology. Physical Chemistry is a significant part of the JEE chemistry syllabus. Within JEE Physical Chemistry, entropy change calculations are a recurrent topic, which blend the fundamental concepts of thermodynamics with their application to chemical processes.

As a JEE aspirant, it's essential to have a strong grasp on topics like entropy, as the exam often includes problems that test students' understanding of heat transfer, energy changes, and spontaneity of reactions. The numerical approach to solving entropy-related problems requires rigorous practice and a firm understanding of the underlying principles to perform well in this competitive exam.

Entropy is a measure of the disorder or randomness within a thermodynamic system. Temperature, on the other hand, relates to the average kinetic energy of particles in a system. As temperature increases, particles move faster and the disorder within the system typically increases, resulting in an increase in entropy. This relationship is an important aspect of the second law of thermodynamics.

The formula for calculating the change in entropy when heat capacity is constant, such as in the textbook exercise, is \(\Delta S = m \times Cp,m \times \ln(T2 / T1)\), where \(m\) is the mass of the substance, \(Cp,m\) is its heat capacity at constant pressure per mole, and \(T1\) and \(T2\) are the initial and final temperatures, respectively. It's crucial to convert all temperatures to Kelvin and carefully apply the formula to determine the entropy change accurately.