/*! 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 29 (a) When ice floats in water at ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) When ice floats in water at \(0^{\circ} \mathrm{C},\) what percent of its volume is submerged? (b) What is the specific gravity of ice?

Short Answer

Expert verified
b) What is the specific gravity of ice? a) The percentage of ice volume submerged in water at 0°C is approximately 109.03%. b) The specific gravity of ice is approximately 0.917.

Step by step solution

01

Recall Archimedes' Principle

According to Archimedes' principle, when an object is immersed in a liquid, the buoyant force acting on the object is equal to the weight of the liquid displaced by the object. Mathematically, this can be written as: \[F_{B} = \rho_{L} V_{L} g\] Where \(F_{B}\) is the buoyant force, \(\rho_{L}\) is the density of the liquid, \(V_{L}\) is the volume of the liquid displaced, and \(g\) is the acceleration due to gravity.
02

Equate buoyant force with the weight of the ice

When the ice floats in water, the buoyant force is equal to the weight of the ice. Therefore, we can write the equation for the ice as: \[F_{B} = m_{ice}g\] Where \(m_{ice}\) is the mass of the ice. Since \(F_{B} = \rho_{L} V_{L} g\), we can also write the equation as: \[m_{ice}g = \rho_{L} V_{L} g\]
03

Calculate the percentage of ice volume submerged

To find the percentage of ice volume submerged, we first need to find the volume of the ice (\(V_{ice}\)) using the mass and density of ice: \[V_{ice} = \frac{m_{ice}}{\rho_{ice}}\] The percentage of ice volume submerged (\(P_{submerged}\)) can be found using the volume of liquid displaced (\(V_{L}\)) and the volume of the ice (\(V_{ice}\)): \[P_{submerged} = \frac{V_{L}}{V_{ice}} \times 100\] Substitute the mass of the ice equation from Step 2: \[\frac{V_{L}g}{V_{ice}g} = \frac{\rho_{L}}{\rho_{ice}}\] Cancel out the acceleration due to gravity and rearrange for \(P_{submerged}\): \[P_{submerged} = \frac{\rho_{L}}{\rho_{ice}} \times 100\] The density of ice and water at \(0^{\circ} \mathrm{C}\) are \(\rho_{ice} = 917 kg/m^3\) and \(\rho_{L} = 1000 kg/m^3\). Substitute these values: \[P_{submerged} = \frac{1000}{917} \times 100 \approx 109.03\%\]
04

Calculate the specific gravity of ice

The specific gravity of a substance is the ratio of its density to the density of a reference substance, typically water. In this case, the specific gravity of ice can be found using its density and the density of water: \[Specific \; Gravity \; (ice) = \frac{\rho_{ice}}{\rho_{L}}\] Substitute the values of the densities: \[Specific \; Gravity \; (ice) = \frac{917}{1000} \approx 0.917\] #Final Answer# (a) The percentage of ice volume submerged in water at 0°C is approximately 109.03%. (b) The specific gravity of ice is approximately 0.917.

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!

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

A water tower supplies water through the plumbing in a house. A 2.54 -cm- diameter faucet in the house can fill a cylindrical container with a diameter of \(44 \mathrm{cm}\) and a height of \(52 \mathrm{cm}\) in 12 s. How high above the faucet is the top of the water in the tower? (Assume that the diameter of the tower is so large compared to that of the faucet that the water at the top of the tower does not move. \()\)
A sphere of radius \(1.0 \mathrm{cm}\) is dropped into a glass cylinder filled with a viscous liquid. The mass of the sphere is \(12.0 \mathrm{g}\) and the density of the liquid is \(1200 \mathrm{kg} / \mathrm{m}^{3} .\) The sphere reaches a terminal speed of \(0.15 \mathrm{m} / \mathrm{s} .\) What is the viscosity of the liquid?
A dinoflagellate takes 5.0 s to travel 1.0 mm. Approximate a dinoflagellate as a sphere of radius \(35.0 \mu \mathrm{m}\) (ignoring the flagellum). (a) What is the drag force on the dinoflagellate in seawater of viscosity $0.0010 \mathrm{Pa} \cdot \mathrm{s} ?$ (b) What is the power output of the flagellate?
The body of a 90.0 -kg person contains \(0.020 \mathrm{m}^{3}\) of body fat. If the density of fat is \(890 \mathrm{kg} / \mathrm{m}^{3},\) what percentage of the person's body weight is composed of fat?
A fish uses a swim bladder to change its density so it is equal to that of water, enabling it to remain suspended under water. If a fish has an average density of \(1080 \mathrm{kg} / \mathrm{m}^{3}\) and mass \(10.0 \mathrm{g}\) with the bladder completely deflated, to what volume must the fish inflate the swim bladder in order to remain suspended in seawater of density $1060 \mathrm{kg} / \mathrm{m}^{3} ?$
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.