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Hot water and a piece of cold metal come into contact in an isolated container. When the final temperature of the metal and water are identical, is the total energy change in this process (a) zero; (b) negative; (c) positive; (d) not enough information.

Short Answer

Expert verified
(a) The total energy change in this process is zero. This is based on the principle of conservation of energy.

Step by step solution

01

Understanding the concept of thermal equilibrium

When two objects of different temperatures are placed in contact within an isolated system, they will eventually achieve thermal equilibrium. This means that they will share the same temperature, with the hotter object losing heat and transferring it to the cooler one.
02

Applying the principle of conservation of energy

The principle of conservation of energy stipulates that energy can neither be created nor destroyed; it only changes from one form to another. In this case, the energy lost by the hot water will be equal to the energy gained by the cold metal. This is a simple application of the First Law of Thermodynamics.
03

Deduction based on the scenario

Given that the system is isolated and heat exchange is only happening between the hot water and the cold metal, it is valid to say that the total energy change of the system is zero. This is because the amount of energy lost by the hot water is equal to the amount of energy gained by the cold metal.

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Key Concepts

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

Conservation of Energy
At the heart of many scientific queries and real-world problems lies the principle of conservation of energy. This fundamental concept asserts that in any closed or isolated system, energy cannot be created or destroyed. It can only be transformed from one form to another. This principle is central to the laws of physics and is particularly crucial in understanding thermal processes.

When considering the scenario of hot water transferring heat to a cold piece of metal, we must recognize that although the temperatures of the objects change, the total energy within the system remains constant. The energy that manifests as heat in the hot water is reduced as it is transferred to the metal, which correspondingly experiences a rise in heat energy. This transformation of energy from the hot water to the cold metal is a pristine example of the conservation of energy in action.

Understanding this principle helps answer the original exercise question where despite changes in individual temperatures, the overall energy within the isolated system does not change, showcasing the unwavering nature of the conservation of energy.
Thermodynamics
Thermodynamics is the study of heat, energy, and the movement of them within physical systems. It's a branch of physics that encompasses principles governing the relationships between different forms of energy and how energy affects matter. One key aspect of thermodynamics is understanding how energy changes form, such as when thermal energy is transferred between objects.

Regarding our exercise, when the hot water and cold metal equilibrium is reached, the laws of thermodynamics dictate that heat energy will no longer flow between the two. This is because the temperatures are the same, and there is no longer a temperature gradient to drive the heat transfer. This outcome is an elegant demonstration of the zeroth law of thermodynamics, which states that if two systems are in thermal equilibrium with a third system, they are also in thermal equilibrium with each other. In simpler terms, systems that are at the same temperature are said to be in thermal equilibrium.
Heat Transfer
Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy (heat) between physical systems. Heat transfer is classified into various mechanisms, such as conduction, convection, and radiation.

In our scenario, conduction is the key mode of heat transfer, with thermal energy moving from the hot water to the colder metal until temperatures equalize. Understanding the process behind heat transfer not only allows us to comprehend how the thermal equilibrium is achieved in the given exercise but also how efficiency and insulation might play roles in everyday applications. From designing better cookware to improving building insulation, the principles of heat transfer are applied ubiquitously in engineering and technology.

To facilitate learning, one useful approach is to visualize heat as a substance that flows—much like water might—from areas of higher concentration (higher temperature) to areas of lower concentration (lower temperature). This 'thermal flow' continues until a balance is reached, and we have thermal equilibrium, which is central to the exercise we are reviewing.

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Most popular questions from this chapter

A \(1.00 \mathrm{g}\) sample of \(\mathrm{Ne}(\mathrm{g})\) at 1 atm pressure and \(27^{\circ} \mathrm{C}\) is allowed to expand into an evacuated vessel of \(2.50 \mathrm{L}\) volume. Does the gas do work? Explain.

The following substances undergo complete combustion in a bomb calorimeter. The calorimeter assembly has a heat capacity of \(5.136 \mathrm{kJ} /^{\circ} \mathrm{C} .\) In each case, what is the final temperature if the initial water temperature is \(22.43^{\circ} \mathrm{C} ?\) \(\begin{array}{lllll}\text { (a) } 0.3268 & \text { g caffeine, } & \mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{N}_{4} & \text { (heat of }\end{array}\) combustion \(=-1014.2 \mathrm{kcal} / \mathrm{mol} \text { caffeine })\) (b) \(1.35 \mathrm{mL}\) of methyl ethyl ketone, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}(1)\) \(d=0.805 \mathrm{g} / \mathrm{mL}\) (heat of combustion \(=-2444 \mathrm{kJ} / \mathrm{mol}\) methyl ethyl ketone).

A 0.205 g pellet of potassium hydroxide, \(\mathrm{KOH}\), is added to \(55.9 \mathrm{g}\) water in a Styrofoam coffee cup. The water temperature rises from 23.5 to \(24.4^{\circ} \mathrm{C}\). [Assume that the specific heat of dilute \(\mathrm{KOH}(aq)\) is the same as that of water.] (a) What is the approximate heat of solution of \(\mathrm{KOH}\) expressed as kilojoules per mole of \(\mathrm{KOH}?\) (b) How could the precision of this measurement be improved without modifying the apparatus?

We can determine the purity of solid materials by using calorimetry. A gold ring (for pure gold, specific heat \(=0.1291 \mathrm{Jg}^{-1} \mathrm{K}^{-1}\) ) with mass of \(10.5 \mathrm{g}\) is heated to \(78.3^{\circ} \mathrm{C}\) and immersed in \(50.0 \mathrm{g}\) of \(23.7^{\circ} \mathrm{C}\) water in a constant-pressure calorimeter. The final temperature of the water is \(31.0^{\circ} \mathrm{C}\). Is this a pure sample of gold?

For the reaction \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(1)\) determine \(\Delta H^{\circ},\) given that $$\begin{array}{r} 4 \mathrm{HCl}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ}=-202.4 \mathrm{kJ} \end{array}$$ $$\begin{aligned} 2 \mathrm{HCl}(\mathrm{g})+\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \\ \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(1)+\mathrm{H}_{2} \mathrm{O}(1) & \Delta H^{\circ}=-318.7 \mathrm{kJ} \end{aligned}$$

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