/*! 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 -kg piece of aluminum th... [FREE SOLUTION] | 91Ó°ÊÓ

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A 0.200 -kg piece of aluminum that has a temperature of \(-155^{\circ} {C}\) is added to 1.5 \({kg}\) of water that has a temperature of \(3.0^{\circ} {C}\) . At equilibrium the temperature is \(0.0^{\circ} {C}\) . 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
Approximately 0.027 kg of water is frozen into ice.

Step by step solution

01

Calculate Heat Loss by Aluminum

First, calculate the heat lost by the piece of aluminum as it reaches the equilibrium temperature. Use the formula \( Q = mc\Delta T \), where \( m \) is the mass of aluminum (0.200 kg), \( c \) is the specific heat capacity of aluminum \( (0.897 \, \text{J/g} \cdot \degree C) \), and \( \Delta T \) is the change in temperature \( (-155^{\circ} \text{C} - 0^{\circ} \text{C}) \). The heat lost by aluminum is given by:\[ Q_{\text{Al}} = 0.200 \, \text{kg} \times 0.897 \, \text{kJ/kg} \cdot \degree C \times (0 - (-155)) \]Convert the specific heat to consistent units: \( 0.897 \, \text{kJ/kg} \cdot \degree C = 897 \, \text{J/kg} \cdot \degree C \). Thus,\[ Q_{\text{Al}} = 0.200 \times 897 \times 155 = 27819 \, \text{J} \]
02

Calculate Heat Gained by Water

Convert the liquid water at 3.0°C to 0°C. Use the same heat equation \( Q = mc\Delta T \). Here, \( m = 1.5 \, \text{kg} \), \( c = 4.18 \, \text{kJ/kg} \cdot \degree C \), converting to \( 4180 \, \text{J/kg} \cdot \degree C \) for consistency, and \( \Delta T = 0^{\circ}C - 3^{\circ}C \).\[ Q_{\text{H2O, heating}} = 1.5 \times 4180 \times 3 = 18810 \, \text{J} \]
03

Calculate Heat Required to Freeze Water

The heat released when water freezes is given by \( Q = mL_f \), where \( L_f = 334,000 \, \text{J/kg} \) is the latent heat of fusion for ice. Let the mass of ice formed be \( m_f \). We need to find this \( m_f \) using the equation:\[ Q_{\text{frozen water}} = m_f \times 334,000 \]
04

Apply Conservation of Energy

Set the heat lost by aluminum equal to the total heat gained by water. Some of it heats water to 0°C, and the rest freezes part of it. This becomes:\[ 27,819 = 18,810 + m_f \times 334,000 \]Solve for \( m_f \):\[ 27919 - 18810 = m_f \times 334,000 \]\[ 9109 = m_f \times 334,000 \]\[ m_f = \frac{9109}{334,000} \approx 0.027 \, \text{kg} \]
05

Conclusion

Calculate the precise mass of water that has been frozen using the conservation of energy calculations. The finalized mathematical result tells us that approximately 0.027 kg of water has been frozen into ice.

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Key Concepts

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

Heat Exchange
Heat exchange is a fundamental concept in thermodynamics that refers to the process where thermal energy is transferred from one object or substance to another. This transfer occurs due to a temperature difference between the two entities. In the context of the exercise, the piece of aluminum and the water exchange heat until they reach a thermal equilibrium.
For heat exchange calculations, it is crucial to apply the specific heat equation. This equation is denoted as \( Q = mc\Delta T \), where \( Q \) is the heat exchanged, \( m \) is the mass of the substance, \( c \) is its specific heat capacity, and \( \Delta T \) is the change in temperature.
In the exercise, the heat lost by the aluminum is equivalent to the heat gained by the water, assuming no heat is lost to the environment. This conservation principle guides us in understanding the energy relationship within the system.
Latent Heat of Fusion
Latent heat of fusion is the amount of energy required to convert a unit mass of a solid into a liquid at a constant temperature and pressure. This energy is absorbed by the substance either from the surroundings or another element in the system. In our example, it involves turning water at 0°C into ice, which is a form of releasing latent heat.
The formula used to calculate the heat exchange during phase change is \( Q = mL_f \), where \( L_f \) is the latent heat of fusion. For water, \( L_f \) is a substantial 334,000 J/kg. This quantity indicates the considerable energy involved to alter its phase from liquid to solid.
Here, the water released heat as it transformed into ice, absorbing energy provided by the aluminum that cooled down.
Specific Heat Capacity of Materials
Specific heat capacity is defined as the heat required to change the temperature of a unit mass of a substance by one-degree Celsius. Each material has a distinct specific heat capacity which determines how much heat it can store.
In the problem, aluminum possesses a specific heat capacity of 897 J/kg⋅°C. This relatively high value means it can store substantial amounts of heat. Conversely, water's specific heat capacity is 4,180 J/kg⋅°C, which is considerably high, allowing it to absorb significant heat before undergoing a temperature change.
  • This property makes water an excellent moderator of thermal energy and is the reason for its widespread use in climate regulation.
  • The heat capacity indicates the energy exchange process in thermal systems, impacting how quickly substances heat up or cool down.
Conservation of Energy
The conservation of energy principle states that energy cannot be created or destroyed, only transformed or transferred. This law is vital in analyzing the heat exchange process in thermodynamic systems.
In the solution to our exercise, this principle was crucial. It helped us equate the heat lost by aluminum to the heat gained by water and the energy needed for freezing, ensuring no heat was lost overall. The mathematical equation for energy conservation in this context was set up as:
\[ Q_{\text{lost by Al}} = Q_{\text{gained by water}} + Q_{\text{energy for freezing}} \]
  • Through this equation, we determine how much of the water changes state from liquid to solid due to the thermal energy exchanged with aluminum.
  • This approach ensures all calculations abide by the universal law of energy conservation.

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Most popular questions from this chapter

A person eats a container of strawberry yogurt. The Nutritional Facts label states that it contains 240 Calories (1 Calorie 4186 J). What mass of perspiration would one have to lose to get rid of this energy? At body temperature, the latent heat of vaporization of water is \(2.42 \times 10^{6} {J} / {kg}\)

An aluminum baseball bat has a length of 0.86 \({m}\) at a temperature of \(17^{\circ} {C}\) . When the temperature of the bat is raised, the bat lengthens by 0.00016 \({m} .\) Determine the final temperature of the bat.

SSM Suppose you are hiking down the Grand Canyon. At the top, the temperature early in the morning is a cool 3 C. By late afternoon, the temperature at the bottom of the canyon has warmed to a sweltering 34 C. What is the difference between the higher and lower temperatures in (a) Fahrenheit degrees and (b) kelvins?

An insulated container is partly filled with oil. The lid of the container is removed, 0.125 kg of water heated to \(90.0^{\circ} {C}\) is poured in, and the lid is replaced. As the water and the oil reach equilibrium, the volume of the oil increases by \(1.20 \times 10^{-5} {m}^{3}\) . The density of the oil is 924 \({kg} / {m}^{3}\) , its specific heat capacity is \(1970{J} /({kg} \cdot {C}^{\circ}),\) and its coefficient of volume expansion is \(721 \times 10^{-6}({C}^{0})^{-1}.\) What is the temperature when the oil and the water reach equilibrium?

Two grams of liquid water are at \(0^{\circ} {C},\) and another two grams are at \(100^{\circ} {C}\) . Heat is removed from the water at \(0^{\circ} {C},\) completely freezing it at 0 'C. This heat is then used to vaporize some of the water at \(100^{\circ} {C}\) . What is the mass (in grams) of the liquid water that remains?

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