/*! 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 68 A 155-g aluminum cylinder is rem... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 155-g aluminum cylinder is removed from a liquid nitrogen bath, where it has been cooled to \(-196^{\circ} \mathrm{C}\). The cylinder is immediately placed in an insulated cup containing 80.0 g of water at \(15.0^{\circ} \mathrm{C}\). What is the equilibrium temperature of this system? If your answer is \(0{ }^{\circ} \mathrm{C}\), determine the amount of water that has frozen. The average specific heat of aluminum over this temperature range is \(653 \mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})\).

Short Answer

Expert verified
Equilibrium temp is 0°C, and 44.4 g of water freezes.

Step by step solution

01

List Given Data

Begin by organizing the given information for easy reference. - Mass of aluminum ( m_a ) = 155 g = 0.155 kg - Initial temperature of aluminum ( T_{a1} ) = -196°C - Specific heat of aluminum ( c_a ) = 653 J/(kg·K) - Mass of water ( m_w ) = 80 g = 0.080 kg - Initial temperature of water ( T_{w1} ) = 15°C - Specific heat of water ( c_w ) = 4186 J/(kg·K) - Fusion temperature of water = 0°C - Heat of fusion of water ( L_f ) = 334,000 J/kg.
02

Set Up Energy Exchange Equation

The heat lost by aluminum will equal the heat gained by water. Use the formula for heat exchange: \( Q = mc\Delta T \).For aluminum: \[ Q_{aluminum} = m_a c_a (T_f - T_{a1}) \]For water:\[ Q_{water} = m_w c_w (T_f - T_{w1}) \]Assuming no phase change occurs, \( Q_{aluminum} + Q_{water} = 0 \). Solve for \( T_f \), the final equilibrium temperature.
03

Calculate Heat Exchange

First, find the heat exchange separately:\[ Q_{aluminum} = 0.155 \times 653 \times (T_f + 196) \]\[ Q_{water} = 0.080 \times 4186 \times (T_f - 15) \]Setting these equations equal: \[ 0.155 \times 653 \times (T_f + 196) + 0.080 \times 4186 \times (T_f - 15) = 0 \]Solve this equation for \( T_f \).
04

Solve for Final Temperature

Solve the equation:\[ 101.215T_f + 19838.14 + 334.88T_f - 5022 = 0 \]Combine like terms:\[ 436.095T_f + 14816.14 = 0 \]\[ 436.095T_f = -14816.14 \]\[ T_f = \frac{-14816.14}{436.095} \approx -33.97°C \]This indicates significant heat exchange which could result in a phase change.
05

Determine if Temperature Reaches 0°C

Because the calculated \( T_f = -33.97°C \) is below 0°C, it's crucial to consider the potential freezing of water. Analyze the energy required to bring the whole system to 0°C and check if water freezes.
06

Calculate if Water Freezes

Calculate the heat required to bring the aluminum to 0°C:\[ Q_{aluminum, to 0} = 0.155 \times 653 \times (0 + 196) = 19838.14 \, ext{J} \]Heat loss for water to cool to 0°C:\[ Q_{water, to 0} = 0.080 \times 4186 \times (0 - 15) = -5020.8 \, ext{J} \]The surplus energy available is \( \approx 14817.34 \, J \).
07

Calculate Amount of Water that Freezes

Calculate the amount of water that freezes using latent heat:\[ Q_{freezing} = m_{ice}L_f \]\[ 14817.34 = m_{ice} \times 334,000 \]\[ m_{ice} = \frac{14817.34}{334,000} \approx 0.0444 \, ext{kg} \]Approximately 44.4 g of water freezes.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Specific Heat Capacity
Specific heat capacity is an essential concept when studying thermal equilibrium. It is a measure of how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius or one Kelvin. Each material has its specific heat capacity, which determines how it reacts to heat changes. For instance, in our example, aluminum has a specific heat capacity of 653 J/(kg·K), which means it requires 653 joules of energy to raise one kilogram of aluminum by one degree Kelvin.
Understanding specific heat capacity helps predict temperature changes when heat energy is exchanged between substances. Substances with a low specific heat capacity heat up or cool down more quickly than those with a higher specific heat capacity. This principle is critical to finding the equilibrium temperature in experiments such as the one with the aluminum cylinder being placed in water. By knowing the specific heat capacities, students can calculate the temperature change of both aluminum and water as they seek thermal equilibrium.
Heat Exchange
Heat exchange refers to the transfer of thermal energy between substances until they reach the same temperature, reaching a state of thermal equilibrium. In our problem, the aluminum cylinder and water exchange heat with one another in an insulated environment.
The heat lost by the aluminum when it warms up is numerically equal to the heat gained by the water as it cools down. This principle is known as the conservation of energy in a closed system.
  • Aluminum releases heat as it rises from -196°C to a higher temperature.
  • Water absorbs the heat, starting at 15°C and moving towards a lower equilibrium temperature.
We calculate this heat exchange using the equation \( Q = mc\Delta T \) for both materials, where \( Q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change. These calculations will help us determine the new equilibrium temperature and whether the water will undergo a phase change.
Phase Change
A phase change occurs when a substance transitions from one state of matter to another, such as from liquid to solid or vice versa. In the provided scenario, when calculating the equilibrium temperature, we noted that the initial theoretical temperature was below 0°C. This indicates that a phase change could take place, specifically the freezing of water.
When the temperature of the system is theoretically below 0°C, water in the system might freeze. To address this, one must calculate the required energy for such a phase transformation. If there's enough energy removed from the water to lower its temperature to the freezing point and continue beyond, a portion of the water will freeze.
Understanding phase changes is crucial, as it significantly affects the final calculations and the amount of thermal energy involved. It's essential to check if the phase boundary—like the freezing point—plays a role in energy exchange calculations.
Latent Heat
Latent heat is the heat required to change the phase of a substance without affecting its temperature. In our example, it specifically refers to the heat needed to freeze the water at 0°C, given by the latent heat of fusion of water, which is 334,000 J/kg.
Once the water's temperature reaches 0°C and further cooling would result in freezing, the energy removed is used to convert liquid water into ice without changing temperature. This energy is called latent heat.
Using the latent heat of fusion formula \( Q = mL_f \), where \( m \) is the mass of the substance changing phase and \( L_f \) is the latent heat of fusion, we can calculate the amount of water that freezes. In our scenario, with about 14817.34 J of surplus energy, around 44.4 g of the water freezes. This process explains the energy distribution when moving beyond mere temperature changes to involve phase transformations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Standard temperature and pressure (STP) is defined as a temperature of \(0^{\circ} \mathrm{C}\) and a pressure of \(101.3 \mathrm{kPa}\). What is the volume occupied by one mole of an ideal gas at STP?

What is the temperature of a gas of \(\mathrm{CO}_{2}\) molecules whose rms speed is \(329 \mathrm{m} / \mathrm{s}\) ?

A compressed-air tank holds \(0.500 \mathrm{m}^{3}\) of air at a temperature of \(285 \mathrm{K}\) and a pressure of \(880 \mathrm{kPa}\). What volume would the air occupy if it were released into the atmosphere, where the pressure is \(101 \mathrm{kPa}\) and the temperature is \(303 \mathrm{K} ?\)

Peter catches a \(4.8-\mathrm{kg}\) striped bass on a fishing line \(0.54 \mathrm{mm}\) in diameter and begins to reel it in. He fishes from a pier well above the water, and his fish hangs vertically from the line out of the water. The fishing line has a Young's modulus of \(5.1 \times 10^{9} \mathrm{N} / \mathrm{m}^{2} .\) (a) What is the fractional increase in length of the fishing line if the fish is at rest? (b) What is the fractional increase in the fishing line's length when the fish is pulled upward with a constant speed of \(1.5 \mathrm{m} / \mathrm{s} ?\) (c) What is the fractional increase in the fishing line's length when the fish is pulled upward with a constant acceleration of \(1.5 \mathrm{m} / \mathrm{s}^{2} ?\)

One mole of a monatomic ideal gas has an initial pressure of \(210 \mathrm{kPa},\) an initial volume of \(1.2 \times 10^{-3} \mathrm{m}^{3}\), and an initial temperature of \(350 \mathrm{K}\). The gas now undergoes three separate processes: (i) a constant-temperature expansion that triples its volume; (ii) a constant-pressure compression to its initial volume; and (iii) a constant-volume increase in pressure to its initial pressure. At the end of these three processes, the gas is back at its initial pressure, volume, and temperature. Plot these processes on a pressure-versus-volume graph, showing the values of \(P\) and \(V\) at the end points of each process.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.