When freezing of a liquid takes place in a system : (a) may have \(q>0\) or \(q<0\) depending on the liquid (b) is represented by \(q>0\) (c) is represented by \(q<0\) (d) has \(q=0\)

Short Answer

Expert verified
The correct response is (c) is represented by \(q<0\), as the liquid releases heat when it freezes.

Step by step solution

01

Understanding Heat Transfer during Phase Change

In thermodynamics, 'q' represents the amount of heat energy transferred. When a liquid freezes, it changes from a liquid to a solid phase. During this process, the liquid releases heat energy to the surrounding environment; hence, the heat energy is negative because the system (liquid) is losing heat.
02

Interpreting the Sign of Heat Transfer

Therefore, for the freezing process the value of 'q' should reflect the loss of heat energy. A negative 'q' signifies that the system is releasing energy, which matches what we expect during freezing.
03

Choosing the Correct Answer

Since the system loses heat when a liquid freezes, the correct representation is 'q<0', meaning that heat is being removed from the liquid during the freezing process.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer
Understanding the mechanisms of heat transfer is central to grasping how energy moves within thermodynamic systems. When a substance undergoes a phase change, like freezing, heat energy is either absorbed or released. This transfer of energy can be quantified by the variable 'q'.

Different materials exhibit unique heat transfer characteristics. For instance, water releases heat as it freezes, whereas some substances, under specific conditions, might absorb heat during a phase transition.

Heat transfers through three primary modes: conduction, convection, and radiation. During the freezing process, conduction is usually the most significant, where heat energy is transferred from the warmer liquid to the colder surrounding environment, leading to the crystallization of the substance into a solid. Of note, the heat transfer during phase changes, including freezing, occurs at a constant temperature, reflecting the unique property of latent heat.
Freezing Process
The freezing process of a liquid into a solid is a fascinating thermodynamic phenomenon that is part of everyday experiences like ice formation. When a liquid freezes, it undergoes a phase change where the substance changes its state from liquid to solid.

At the freezing point, the molecules within the liquid slow down due to the decrement in their thermal energy. As energy is transferred out of the system, the temperature remains constant despite the ongoing heat loss. This equilibrium maintains until the entire liquid is transformed into a solid. The transformation results in an energy discharge in the form of heat, known as the latent heat of fusion, and is identified by a negative 'q' value, stating that the system is losing heat.

The efficiency of this process is influenced by various factors including pressure, purity of the substance, and the presence of impurities or nucleating agents which can alter the freezing point.
Thermodynamic Systems
In understanding thermodynamic systems, it's important to define the system and its surroundings. The system is usually the substance or mixture of substances undergoing a thermodynamic process, like freezing, and everything outside of this area is considered the surroundings.

Systems can be isolated (no heat or matter exchange), closed (only energy exchange), or open (exchange of both energy and matter). During the freezing process in a closed system, the substance (system) releases heat to its surroundings (environment).

The study of these systems and how they interact with the environment is integral to thermodynamics. Whether discussing engines, refrigerators, or biological processes, the laws governing energy transformations remain the same. The first law of thermodynamics, for example, focuses on the conservation of energy, ensuring that all heat lost by the system must be accounted for in its surroundings.

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Most popular questions from this chapter

If one mole of an ideal gas \(\left(C_{p, m}=\frac{5}{2} R\right)\) is expanded isothermally at 300 until it's volume is tripled, then change in entropy of gas is : (a) zero (b) infinity (c) \(\frac{5}{2} R \ln 3\) (d) \(R \ln 3\)

Using the listed \(\left[\Delta G^{\circ}{ }_{f}\right.\) values \(]\) calculate \(\Delta G^{\circ}\) for the reaction : \(3 \mathrm{H}_{2} \mathrm{~S}(g)[-33.6]+2 \mathrm{HNO}_{3}(l)[-80.6] \stackrel 2 \mathrm{NO}(g)[+86.6]+4 \mathrm{H}_{2} \mathrm{O}(l)[-237.1]+3 S(s)[0.0]\) (a) \(-513.0\) (b) \(-1037.0\) (c) \(+433.4\) (d) \(+225.0\)

A \(0.05 \mathrm{~L}\) sample of \(0.2 \mathrm{M}\) aqueous hydrochloric acid is added to \(0.05 \mathrm{~L}\) of \(0.2 \mathrm{M}\) aqueous ammonia in a calorimeter. Heat capacity of entire calorimeter system is \(480 \mathrm{~J} / \mathrm{K}\). The temperature increase is \(1.09 \mathrm{~K}\). Calculate \(\Delta_{r} H^{\circ}\) in \(\mathrm{kJ} / \mathrm{mol}\) for the following reaction: $$ \mathrm{HCl}(a q .)+\mathrm{NH}_{3}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ (a) \(-52.32\) (b) \(-61.1\) (c) \(-55.8\) (d) \(-58.2\)

Calculate the entropy change \((\mathrm{J} / \mathrm{mol} \mathrm{K})\) of the given reaction. The molar entropies [J/K-mol] are given in brackets after each substance. \(2 \mathrm{PbS}(s)[91.2]+3 \mathrm{O}_{2}(g)[205.1] \longrightarrow 2 \mathrm{PbO}(s)[66.5]+2 \mathrm{SO}_{2}(g)[248.2]\) (a) \(-113.5\) (b) \(-168.3\) (c) \(+72.5\) (d) \(-149.2\)

Calculate \(\Delta S\) for following process: $$ \underset{\text { at } 100 \mathrm{~K}}{X(s)} \longrightarrow \underset{\text { at } 200 \mathrm{~K}}{X(l)} $$ Given : Melting point of \(X_{(s)}=100 \mathrm{~K} ; \Delta H_{\text {Fusion }}=20 \mathrm{~kJ} / \mathrm{mol} ; C_{p, m}(X, l)=10 \mathrm{~J} / \mathrm{mol} \mathrm{K}\) (a) \(26.93 \mathrm{~J} / \mathrm{K}\) (b) \(206.93 \mathrm{~J} / \mathrm{K}\) (c) \(203 \mathrm{~J} / \mathrm{K}\) (d) \(206.93 \mathrm{~kJ} / \mathrm{K}\)

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