/*! 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 26 (II) A 58 -kg ice-skater moving ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(II) A 58 -kg ice-skater moving at 7.5 \(\mathrm{m} / \mathrm{s}\) glides to a stop. Assuming the ice is at \(0^{\circ} \mathrm{C}\) and that 50\(\%\) of the heat generated by friction is absorbed by the ice, how much ice melts?

Short Answer

Expert verified
About 0.15 kg of ice melts.

Step by step solution

01

Calculate Kinetic Energy of Skater

The kinetic energy (KE) of the skater can be calculated using the formula: \( KE = \frac{1}{2}mv^2 \) where \( m = 58 \text{ kg} \) and \( v = 7.5 \text{ m/s} \). So, \( KE = \frac{1}{2} \times 58 \times (7.5)^2 \).
02

Calculate Heat Generated

Since all the kinetic energy is converted to heat due to friction to stop the skater, the heat generated is equal to the initial kinetic energy. Hence, \( Q_{\text{total}} = KE \).
03

Determine Heat Absorbed by Ice

From the problem, 50\( \% \) of the heat generated is absorbed by ice. Therefore, the heat absorbed by the ice can be calculated as \( Q_{\text{ice}} = 0.5 \times KE \).
04

Calculate Mass of Ice Melted

The latent heat of fusion of ice is \( L = 334,000 \text{ J/kg} \). The mass of ice melted (\( m \)) can be found using the formula: \( Q_{\text{ice}} = m \times L \). Rearranging gives \( m = \frac{Q_{\text{ice}}}{L} \). Substitute \( Q_{\text{ice}} \) from the previous step to find \( m \).

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.

Heat Generation
When objects move, they possess energy called kinetic energy (KE). In the case of the ice-skater, this energy is transferred into heat energy due to friction when they come to a stop. To compute this heat energy, we first calculate the skater's kinetic energy using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the skater and \( v \) is their velocity. Friction converts all of this kinetic energy into heat (\( Q_{\text{total}} = KE \)).
However, not all the generated heat contributes to melting the ice; only a portion of it is absorbed by the ice. In the exercise, it's specified that 50\(\%\) of the generated heat is absorbed by the ice, defining the equation \( Q_{\text{ice}} = 0.5 \times KE \). This absorbed heat then plays its part in the subsequent melting of the ice.
Latent Heat of Fusion
The latent heat of fusion is a key concept when discussing phase changes. It's the amount of heat energy required to change a substance from solid to liquid without a change in temperature. For ice, this value is given as \( 334,000 \text{ J/kg} \).
In this exercise, once the heat energy is absorbed by the ice due to friction, it doesn't immediately heat up the ice. Instead, it uses that energy to melt, signifying a phase change from solid to liquid. This happens at a constant temperature, as indicated by the concept of latent heat. Understanding this concept is crucial in predicting how much ice will melt based on the absorbed energy. The relationship is described by the equation \( Q_{\text{ice}} = m \times L \), where \( m \) is the mass of the ice melted.
Ice Melting Calculation
The final part of the exercise involves calculating how much ice actually melts. Using the absorbed heat \( Q_{\text{ice}} \) calculated from earlier, we apply the formula for latent heat: \( Q_{\text{ice}} = m \times L \). Here, \( L \) is the latent heat of fusion, a constant \( 334,000 \text{ J/kg} \), and \( m \) is the mass of ice melted.
To find the mass \( m \), we rearrange the equation to \( m = \frac{Q_{\text{ice}}}{L} \). This formula helps determine the quantity of ice that transforms from solid to liquid using the portion of heat it absorbs. Breaking down problems like this step-by-step instills a clear understanding of physical processes and how energy transfers function during them.

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

(I) One end of a 45 -cm-long copper rod with a diameter of 2.0 \(\mathrm{cm}\) is kept at \(460^{\circ} \mathrm{C},\) and the other is immersed in water at \(22^{\circ} \mathrm{C}\) Calculate the heat conduction rate along the rod.

A diesel engine accomplishes ignition without a spark plug by an adiabatic compression of air to a temperature above the ignition temperature of the diesel fuel, which is injected into the cylinder at the peak of the compression. Suppose air is taken into the cylinder at \(280 \mathrm{~K}\) and volume \(V_{1}\) and is compressed adiabatically to \(560^{\circ} \mathrm{C}\left(\approx 1000^{\circ} \mathrm{F}\right)\) and volume \(V_{2}\) Assuming that the air behaves as an ideal gas whose ratio of \(C_{P}\) to \(C_{V}\) is \(1.4,\) calculate the compression ratio \(V_{1} / V_{2}\) of the engine.

A house has well-insulated walls \(19.5 \mathrm{~cm}\) thick (assume conductivity of air) and area \(410 \mathrm{~m}^{2},\) a roof of wood \(5.5 \mathrm{~cm}\) thick and area \(280 \mathrm{~m}^{2}\), and uncovered windows \(0.65 \mathrm{~cm}\) thick and total area \(33 \mathrm{~m}^{2} .\) ( \(a\) ) Assuming that heat is lost only by conduction, calculate the rate at which heat must be supplied to this house to maintain its inside temperature at \(23^{\circ} \mathrm{C}\) if the outside temperature is \(-15^{\circ} \mathrm{C}\). \((b)\) If the house is initially at \(12^{\circ} \mathrm{C}\), estimate how much heat must be supplied to raise the temperature to \(23^{\circ} \mathrm{C}\) within 30 min. Assume that only the air needs to be heated and that its volume is \(750 \mathrm{~m}^{3}\). ( \(c\) ) If natural gas costs \(\$ 0.080\) per kilogram and its heat of combustion is \(5.4 \times 10^{7} \mathrm{~J} / \mathrm{kg},\) how much is the monthly cost to maintain the house as in part \((a)\) for \(24 \mathrm{~h}\) each day, assuming \(90 \%\) of the heat produced is used to heat the house? Take the specific heat of air to be \(0.24 \mathrm{kcal} / \mathrm{kg} \cdot \mathrm{C}^{\circ}\)

(II) One and one-half moles of an ideal monatomic gas expand adiabatically, performing \(7500 \mathrm{~J}\) of work in the process. What is the change in temperature of the gas during this expansion?

(II) Determine \((a)\) the work done and \((b)\) the change in internal energy of 1.00 \(\mathrm{kg}\) of water when it is all boiled to steam at \(100^{\circ} \mathrm{C}\) Assume a constant pressure of 1.00 atm.

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.