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A phase transition occurs in a sample of an alloy, and 437 kJ transfers from the surroundings to the alloy. (Assume that no work is done.) (a) What is the algebraic sign of \(\Delta T_{\text {alloy }}\) ? (b) What is the algebraic sign of \(\Delta_{\mathrm{r}} E_{\text {alloy }} ?\)

Short Answer

Expert verified
(a) Zero; (b) Positive.

Step by step solution

01

Understanding Heat Transfer

In the problem, 437 kJ of heat is transferred from the surroundings to the alloy. This means the alloy absorbs heat.
02

Determine Sign of Temperature Change

When an object absorbs heat, its temperature typically increases unless it is at a phase transition where the temperature remains constant. In this case, since it's a phase transition, the temperature does not change, so \( \Delta T_{\text{alloy}} = 0 \). Thus, the algebraic sign is zero.
03

Determine Sign of Change in Internal Energy

According to the first law of thermodynamics, the change in internal energy \( \Delta E \) is equal to the heat added to the system minus the work done by the system. Here, since 437 kJ of heat is added and no work is done, \( \Delta E = 437 \text{ kJ} \). Therefore, the algebraic sign of \( \Delta_{\mathrm{r}} E_{\text {alloy}} \) is positive.

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Key Concepts

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

Phase Transition
During a phase transition, a material changes from one state of matter to another, such as from solid to liquid or liquid to gas. This process involves a unique aspect of thermodynamics where, during the transition, the temperature of the material does not change, even though it is absorbing or releasing energy.
In phase transitions:
  • Energy is absorbed or released, but temperature remains constant.
  • The energy goes into changing the state of the substance, not changing its temperature.
  • Examples include melting (solid to liquid) and boiling (liquid to gas).
When an alloy undergoes a phase transition, as in this exercise, it absorbs energy from its surroundings. Although it gains energy, there's no temperature change, hence the temperature change (\( \Delta T_{\text{alloy}} \)) is zero. Understanding this concept helps clarify why heat absorption during phase changes doesn't always lead to temperature change.
First Law of Thermodynamics
The first law of thermodynamics, also known as the law of energy conservation, is a fundamental concept in physics. It states that energy cannot be created or destroyed, only converted from one form to another. The equation representing this principle is:\[ \Delta E = Q - W \]
  • \(\Delta E\) is the change in internal energy of the system.
  • \(Q\) is the heat added to the system.
  • \(W\) is the work done by the system.
In the given problem, the alloy is not doing any work (\(W = 0\)). The only energy exchange occurs through heat, where \(Q = 437 \text{ kJ}\) is transferred to the alloy. Therefore, the change in internal energy of the alloy (\(\Delta E_{\text{alloy}}\)) is directly equal to the heat added, resulting in a positive change. This indicates that the internal energy of the alloy increases by 437 kJ.
Heat Transfer
Heat transfer is the movement of thermal energy from one object or material to another. It can occur through conduction, convection, or radiation. In this specific exercise, the focus is on heat transfer during a phase transition:
  • Conduction involves direct contact and is typical in solids.
  • Convection occurs in fluids where the warmer part moves, transferring heat.
  • Radiation involves heat transfer through electromagnetic waves and can happen in a vacuum.
For the alloy, heat transfer is vital to its phase transition. The system (the alloy) absorbs 437 kJ of thermal energy from its surroundings, allowing it to change phase without a temperature change. Heat absorption is a significant part of phase transitions, facilitating the state change while maintaining equilibrium in energy conservation. When heat is added, but no temperature change occurs, it indicates that the energy is being used to break intermolecular bonds, transitioning the material to a different state.

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Most popular questions from this chapter

Melting lead requires 5.50 cal/g. Calculate how many joules are required to melt \(1.00 \mathrm{lb}\) (454 g) lead.

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