/*! 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 67 A 42-kg block of ice at \(0^{\ci... [FREE SOLUTION] | 91Ó°ÊÓ

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A 42-kg block of ice at \(0^{\circ} \mathrm{C}\) is sliding on a horizontal surface. The initial speed of the ice is \(7.3 \mathrm{m} / \mathrm{s}\) and the final speed is \(3.5 \mathrm{m} / \mathrm{s}\). Assume that the part of the block that melts has a very small mass and that all the heat generated by kinetic friction goes into the block of ice. Determine the mass of ice that melts into water at \(0^{\circ} \mathrm{C}\).

Short Answer

Expert verified
2.58 grams of ice melts due to the kinetic energy lost.

Step by step solution

01

Understanding the Problem

We have a 42-kg block of ice moving on a horizontal surface. Its speed decreases from 7.3 m/s to 3.5 m/s. We need to find out the mass of ice that melts due to the kinetic energy lost as heat overcoming the kinetic friction. The melting occurs at 0°C.
02

Energy Lost Due to Friction

The kinetic energy lost as the block slows down is converted into heat. Calculate the initial kinetic energy \[KE_{initial} = \frac{1}{2} m v_{initial}^2 = \frac{1}{2} \times 42 \times (7.3)^2 \text{ J}\]and the final kinetic energy \[KE_{final} = \frac{1}{2} m v_{final}^2 = \frac{1}{2} \times 42 \times (3.5)^2 \text{ J}\].The energy lost, \(E_{lost}\), is \[E_{lost} = KE_{initial} - KE_{final}\].
03

Calculate the Energy Lost

Compute the difference in initial and final kinetic energies:- Initial kinetic energy, \(KE_{initial} = \frac{1}{2} \times 42 \times 7.3^2 = 1118.49 \text{ J}\)- Final kinetic energy, \(KE_{final} = \frac{1}{2} \times 42 \times 3.5^2 = 257.25 \text{ J}\)Thus, \[E_{lost} = 1118.49 - 257.25 = 861.24 \text{ J}\].
04

Determine the Mass of Ice that Melts

The heat energy from friction causes some of the ice to melt. The latent heat of fusion of ice is \(L_f = 334,000 \text{ J/kg}\). The mass \(m\) of ice that melts can be calculated with \[E_{lost} = m L_f \rightarrow m = \frac{E_{lost}}{L_f} = \frac{861.24}{334,000}\].Calculate \(m\).
05

Calculating the Result

By calculating the mass, we get:\[m = \frac{861.24}{334,000} \approx 0.00258 \text{ kg}\], which is 2.58 grams. This is the amount of ice that melts.

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Key Concepts

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

Latent Heat of Fusion
The latent heat of fusion is a critical concept in understanding phase changes. It refers to the amount of heat energy required to change 1 kg of a substance from solid to liquid without changing its temperature. For ice, this value is quite substantial at about 334,000 J/kg. This means when ice melts, it absorbs energy efficiently without a rise in temperature. In the context of the sliding ice block, the kinetic energy lost due to slowing down is converted into the energy needed to melt a small portion of the ice. Understanding latent heat helps us predict how much of a substance will undergo a phase change when a certain amount of energy is applied. In calculations dealing with energy transformation like the one in our problem, the latent heat of fusion allows us to determine the mass of the ice that will transform into water.
Kinetic Friction
Kinetic friction occurs when two objects are in motion relative to each other. In our scenario, the ice block sliding on the horizontal surface encounters kinetic friction, which acts opposite to its motion. This force is responsible for the ice block slowing down, converting its kinetic energy into heat. Frictional forces depend on the nature of the surfaces in contact and the normal force exerting on the object sliding. For ice, this friction is often low due to its smooth nature but still significant enough to cause a conversion of kinetic energy into thermal energy. In this problem, the heat generated due to kinetic friction contributes directly to the melting of the ice.
Energy Conversion
Energy conversion refers to the process of changing energy from one form to another. In this exercise, the kinetic energy of the moving ice block is converted into thermal energy due to the kinetic friction. This conversion is crucial as it supplies the heat required for the melting process. Initially, the ice block has a certain amount of kinetic energy because of its motion. As it slows down, energy is not lost but transformed into heat energy, which tends to spread along the surfaces in contact. Understanding energy conversion is essential to determine how energy behaves in various systems, enabling us to solve for unknowns like the mass of melted ice.
Melting Process
The melting process is a phase transition where a solid turns into a liquid. For ice, this occurs at 0°C, precisely where kinetic energy loss from friction can provide the necessary heat for the melting process without changing temperature. In our problem, the ice's initial kinetic energy decreases, transforming into thermal energy due to friction. This thermal energy is absorbed by a portion of the ice, facilitating its change from solid to liquid. It should be noted that the mass fraction of ice melting is exceptionally small due to the high energy requirement, encapsulated by the latent heat of fusion. Grasping the melting process is crucial in interpreting how external energy influences physical state changes.

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Most popular questions from this chapter

Ideally, when a thermometer is used to measure the temperature of an object, the temperature of the object itself should not change. However, if a significant amount of heat flows from the object to the thermometer, the temperature will change. A thermometer has a mass of\(31.0 \mathrm{g},\) a specific heat capacity of \(c=815 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right),\) and a temperature of \(12.0^{\circ} \mathrm{C} .\) It is immersed in \(119 \mathrm{g}\) of water, and the final temperature of the water and thermometer is \(41.5^{\circ} \mathrm{C} .\) What was the temperature of the water before the insertion of the thermometer?

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An unknown material has a normal melting/freezing point of \(-25.0^{\circ} \mathrm{C},\) and the liquid phase has a specific heat capacity of \(160 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) One-tenth of a kilogram of the solid at \(-25.0^{\circ} \mathrm{C}\) is put into a \(0.150-\mathrm{kg}\) aluminum calorimeter cup that contains \(0.100 \mathrm{kg}\) of glycerin. The temperature of the cup and the glycerin is initially \(27.0^{\circ} \mathrm{C}\). All the unknown material melts, and the final temperature at equilibrium is \(20.0^{\circ} \mathrm{C} .\) The calorimeter neither loses energy to nor gains energy from the external environment. What is the latent heat of fusion of the unknown material?

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