/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 88 A \(0.200-\mathrm{kg}\) piece of... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(0.200-\mathrm{kg}\) piece of aluminum that has a temperature of \(-155^{\circ} \mathrm{C}\) is added to \(1.5 \mathrm{kg}\) of water that has a temperature of \(3.0^{\circ} \mathrm{C}\). At equilibrium the temperature is \(0.0^{\circ} \mathrm{C}\). Ignoring the container and assuming that the heat exchanged with the surroundings is negligible, determine the mass of water that has been frozen into ice.

Short Answer

Expert verified
At equilibrium, a portion of the water freezes due to the heat transfer from aluminum.

Step by step solution

01

Identify the system and components

This problem involves a piece of aluminum and a quantity of water reaching thermal equilibrium. The initial temperatures of both are given, as well as the final equilibrium temperature. We need to determine how much water freezes when the aluminum heats up and the water cools down.
02

Determine the variables involved

The specific heat of aluminum ( {1stC_s d_e_g er sim b o p[allow dari u=m_alu} ) is given, and also the mass of aluminum and its initial temperature as well as the initial temperature of the water. The heat lost from the aluminum equals the heat gained by the water, a crucial concept for solving this problem.
03

Calculate heat changes for aluminum

For the aluminum: Q_{aluminum = c_aluminum n((t_2- (sim , ( ( b _-aluminum ) _al_{ dari t_con , ) )( (c wiz for if dis.lower) ares i.e-wm (con) alu) _al_{ur('swum) -_{er_war = darinheri_alu) ) _{sider-alu_alu). t in s p sec_p( (c_con steel =where al_alu) alu - type: w {n 1 _ er_wari_pi_alu { '_ual_w =_u' represent _'_atio alu_alu Q_alu = Q_water. For the water: Q_water - Q_alu _the ra_watersa_s.ali_alu, O_a . d" .". wateratioleis. r .

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Key Concepts

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

Specific Heat
Specific heat is a property of a substance that tells us how much energy it takes to raise the temperature of one kilogram of the substance by one degree Celsius. This means different substances can absorb varying amounts of energy for the same change in temperature. For example, aluminum has a specific heat of about 900 J/kg°C. This indicates that it requires 900 joules of energy to increase the temperature of 1 kilogram of aluminum by 1°C.

The concept of specific heat is crucial when analyzing problems involving heat exchange, such as in our exercise. Here, specific heats of the substances give us insight into how much energy will transfer between the aluminum and the water as they reach thermal equilibrium. Knowing the specific heat allows us to calculate the thermal energy change in the aluminum, which initially heats up from colder temperatures.

In practical terms, if we know that aluminum has a lower specific heat compared to water, it will cool down or heat up more quickly than water when the two substances are in contact. That's because it needs less energy per kilogram for the same increase in temperature. This characteristic plays an important role in scenarios where materials with different specific heats need to interact thermally.
Heat Transfer
Heat transfer occurs when thermal energy flows from a hotter substance to a cooler one until they reach thermal equilibrium. This is a fundamental principle in thermodynamics. In our exercise, heat is transferred from the piece of aluminum (which begins at a colder temperature) to the water until both reach the same equilibrium temperature of 0°C.

There are three main mechanisms for heat transfer: conduction, convection, and radiation, with conduction being most relevant here. Conduction is the process where heat flows through a solid or between solid substances in direct contact. The atoms and molecules in the solid vibrate and collide, passing energy to one another. In our scenario, as the aluminum warms up and the water cools down, the energy exchange between them demonstrates heat transfer through conduction.

Understanding the concept of heat transfer helps us grasp that the energy lost by the aluminum is precisely balanced by the energy gained by the water before any of it freezes. This idea is crucial to figuring out how much of the water turns into ice.
Phase Change
A phase change is the transformation of a substance from one state of matter to another, such as from liquid to solid. In our exercise, water undergoes a phase change into ice at 0°C, the freezing point of water. Phase changes occur at constant temperatures, engineered by breaking or forming bonds between particles. This energy doesn't change the substance's temperature but changes its state.

When water freezes to become ice, it releases energy, known as latent heat. This is the energy needed to change water into ice without changing its temperature. In thermodynamics, the latent heat of fusion for water is about 334,000 J/kg. This value tells us how much energy must be removed from water to turn it into ice at the same temperature.

During this exercise, as the aluminum heats up, some water releases energy and freezes. To determine how much water has frozen, we consider the energy change in aluminum and use the latent heat concept. It gives students practical insight into real-world phenomena like freezing water and the nuances of energy management in thermal equilibrium scenarios.

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Most popular questions from this chapter

When the temperature of a coin is raised by \(75 \mathrm{C}^{\circ},\) the coin's diameter increases by \(2.3 \times 10^{-5} \mathrm{m} .\) If the original diameter of the coin is \(1.8 \times 10^{-2} \mathrm{m},\) find the coefficient of linear expansion.

An insulated container is partly filled with oil. The lid of the container is removed, \(0.125 \mathrm{kg}\) of water heated to \(90.0^{\circ} \mathrm{C}\) is poured in, and the lid is replaced. As the water and the oil reach equilibrium, the volume of the oil increases by \(1.20 \times 10^{-5} \mathrm{m}^{3} .\) The density of the oil is \(924 \mathrm{kg} / \mathrm{m}^{3},\) its specific heat capacity is \(1970 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\), and its coefficient of volume expansion is \(721 \times 10^{-6}\left(\mathrm{C}^{\circ}\right)^{-1} .\) What is the temperature when the oil and the water reach equilibrium?

The heating element of a water heater in an apartment building has a maximum power output of \(28 \mathrm{kW}\). Four residents of the building take showers at the same time, and each receives heated water at a volume flow rate of \(14 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s}\). If the water going into the heater has a temperature of \(11{ }^{\circ} \mathrm{C},\) what is the maximum possible temperature of the hot water that each showering resident receives?

When it rains, water vapor in the air condenses into liquid water, and energy is released. (a) How much energy is released when \(0.0254 \mathrm{m}\) (one inch) of rain falls over an area of \(2.59 \times 10^{6} \mathrm{m}^{2}\) (one square mile)? (b) If the average energy needed to heat one home for a year is \(1.50 \times 10^{11}\) J, how many homes could be heated for a year with the energy determined in part (a)?

A piece of glass has a temperature of \(83.0^{\circ} \mathrm{C} .\) Liquid that has a temperature of \(43.0^{\circ} \mathrm{C}\) is poured over the glass, completely covering it, and the temperature at equilibrium is \(53.0^{\circ} \mathrm{C} .\) The mass of the glass and the liquid is the same. Ignoring the container that holds the glass and liquid and assuming that the heat lost to or gained from the surroundings is negligible, determine the specific heat capacity of the liquid.

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