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Chapter 12: Problem 61

Interactive Solution \(12.6112 .61\) at provides a model for solving problems such as this. A \(42-\mathrm{kg}\) 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, and determine the mass of ice that melts into water at \(0{ }^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The mass of ice that melts is approximately 0.21 kg.

Step by step solution

01

Understand the Energy Conversion

The kinetic friction between the ice block and the surface generates heat, which causes some of the ice to melt. Therefore, the initial kinetic energy of the block is partially converted into heat.
02

Calculate the Change in Kinetic Energy

Use the formula for kinetic energy \( KE = \frac{1}{2}mv^2 \). Calculate the initial and final kinetic energy: \[ KE_{initial} = \frac{1}{2} \times 42 \times (7.3)^2 \] \[ KE_{final} = \frac{1}{2} \times 42 \times (3.5)^2 \] The change in kinetic energy, \( \Delta KE \), is \( KE_{initial} - KE_{final} \).
03

Determine the Energy Required to Melt Ice

The energy that causes the ice to melt is equal to \( \Delta KE \). The heat of fusion for ice is \( L_f = 334,000 \text{ J/kg} \). Thus, the mass of ice that melts, \( m \), can be calculated using \[ \Delta KE = m \times L_f \].
04

Solve for Mass of Melted Ice

Rearrange the equation from Step 3: \[ m = \frac{\Delta KE}{L_f} \]. Substitute the calculated \( \Delta KE \) and the heat of fusion into the equation to find the mass, \( m \), of the melted ice.

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Key Concepts

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

Heat of Fusion
The heat of fusion is a critical concept when understanding phase changes, such as the melting of ice. It is the amount of energy needed to change a substance from a solid to a liquid without changing its temperature. For ice, this energy is quite significant because it helps overcome the molecular interactions keeping it in solid form.
In practical terms, the heat of fusion for ice is specified as 334,000 Joules per kilogram (J/kg). This means that in order to melt 1 kilogram of ice at 0掳C, 334,000 Joules of energy are required. This energy goes into breaking the hydrogen bonds between water molecules in the ice, allowing them to shift into a liquid state.
Energy Conversion
Energy conversion is a fundamental principle in physics where one form of energy changes into another. In the context of the given problem, the scenario depicts the conversion of kinetic energy into thermal energy due to friction.
When the ice block slides over a surface, its movement generates kinetic energy, which is initially high due to the block's speed. As the ice block slows down, some of this kinetic energy is converted into heat due to kinetic friction between the ice and the surface. This thermal energy then contributes to melting the ice, changing its physical state from solid to liquid.
Melting of Ice
The melting of ice is a common process that requires the input of energy specifically, thermal energy which is provided by the heat of fusion. When ice melts at its melting point of 0掳C, it undergoes a phase transition from solid to liquid.
In the exercise, as the ice block slides and slows down, the kinetic energy lost is turned into thermal energy, effectively melting a portion of the ice. The small mass of the melted ice shows how energy conservation laws apply and how specific energy values, like the heat of fusion, are crucial in determining how much ice undergoes this transformation.
Kinetic Friction
Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. It plays a critical role in energy conversion within this exercise.
As the ice block moves across the surface, kinetic friction works against it, slowing it down and converting some kinetic energy into heat. The amount of kinetic frictional force depends on factors such as the nature of the surface and the normal force鈥� here, the block's weight.
  • Kinetic friction is a dissipative force, meaning it disperses mechanical energy, usually in the form of heat.
  • This heat is absorbed by the ice, leading to its melting, as detailed in the exercise solution.
  • Understanding kinetic friction helps explain how seemingly mechanical problems involve thermodynamics.

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Most popular questions from this chapter

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

Interactive Solution \(\underline{12.61} 12.61\) at provides a model for solving problems such as this. A \(42-\mathrm{kg}\) 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, and determine the mass of ice that melts into water at \(0^{\circ} \mathrm{C}\).

A \(0.35-\mathrm{kg}\) coffee mug is made from a material that has a specific heat capacity of \(920 \mathrm{~J} /\) \(\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and contains \(0.25 \mathrm{~kg}\) of water. The cup and water are at \(15^{\circ} \mathrm{C}\). To make a cup of coffeee, a small electric heater is immersed in the water and brings it to a boil in three minutes. Assume that the cup and water always have the same temperature and determine the minimum power rating of this heater.

To help prevent frost damage, fruit growers sometimes protect their crop by spraying it with water when overnight temperatures are expected to go below the freezing mark. When the water turns to ice during the night, heat is released into the plants, thereby giving them a measure of protection against the falling temperature. Suppose a grower sprays \(7.2 \mathrm{~kg}\) of water at \(0^{\circ} \mathrm{C}\) onto a fruit tree. (a) How much heat is released by the water when it freezes? (b) How much would the temperature of a \(180-\mathrm{kg}\) tree rise if it absorbed the heat released in part (a)? Assume that the specific heat capacity of the tree is \(2.5 \times 10^{3} \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and that no phase change occurs within the tree itself.

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.
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