/*! 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 50 Two small blackened spheres of i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 50

Two small blackened spheres of identical size-one of aluminum, the other of an unknown alloy of high conductivity-are suspended by thin wires inside a large cavity in a block of melting ice. It is found that it takes \(4.8\) minutes for the temperature of the aluminum sphere to drop from \(3^{\circ} \mathrm{C}\) to \(1^{\circ} \mathrm{C}\), and \(9.6\) minutes for the alloy sphere to undergo the same change. If the specific gravities of the aluminum and alloy are \(2.7\) and \(5.4\), respectively, and the specific heat of the aluminum is \(900 \mathrm{~J} / \mathrm{kg} \mathrm{K}\), what is the specific heat of the alloy?

Short Answer

Expert verified
The specific heat of the alloy is 450 J/kg K.

Step by step solution

01

Understand the Problem

We have two spheres, one made of aluminum and the other of an unknown alloy, both placed inside melting ice. Their temperature drops from \(3^{\circ} \text{C}\) to \(1^{\circ} \text{C}\). We know the time it takes for this change, the specific gravities, and the specific heat of aluminum. We need to find the specific heat capacity of the unknown alloy.
02

Apply the Heat Transfer Formula

For both spheres, the heat lost can be expressed as \( Q = mc\Delta T \) where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the temperature change. The rate of heat transfer, \( \dot{Q} = \frac{Q}{t} = mc\frac{\Delta T}{t} \), is the same for both spheres because they are in identical environments.
03

Set Up the Equation for Aluminum

For the aluminum sphere, use \( m_{\text{Al}}c_{\text{Al}}\frac{\Delta T}{t_1} \). We know \( c_{\text{Al}} = 900 \, \text{J/kg K}\), specific gravity \( s_{\text{Al}} = 2.7 \), and \( t_1 = 4.8 \) minutes. Therefore:\[ m_{\text{Al}} = V \times s_{\text{Al}} = \gamma \times 2.7 \]where \( \gamma \) is a constant (volume multiplied with density is mass). Then substituting in the heat equation gives:\[ \dot{Q} = \gamma \times 2.7 \times 900 \times \frac{2}{t_1} \]
04

Set Up the Equation for the Alloy

For the alloy sphere, use a similar approach: \( m_{\text{Alloy}}c_{\text{Alloy}}\frac{\Delta T}{t_2} \). The specific gravity \( s_{\text{Alloy}} = 5.4 \) and \( t_2 = 9.6 \) minutes:\[ m_{\text{Alloy}} = V \times s_{\text{Alloy}} = \gamma \times 5.4 \]Substituting into the heat equation gives:\[ \dot{Q} = \gamma \times 5.4 \times c_{\text{Alloy}} \times \frac{2}{t_2} \]
05

Equate the Heat Transfer Rates

Since the rate of heat transfer \( \dot{Q} \) is the same for both spheres, equate the expressions obtained:\[ \gamma \times 2.7 \times 900 \times \frac{2}{4.8} = \gamma \times 5.4 \times c_{\text{Alloy}} \times \frac{2}{9.6} \]The \( \gamma \) and \(\frac{2}{\Delta T}\) cancel out, simplifying to:\[ \frac{2.7 \times 900}{4.8} = \frac{5.4 \times c_{\text{Alloy}}}{9.6} \]
06

Solve for Specific Heat of the Alloy

Rearrange to find \( c_{\text{Alloy}} \):\[ c_{\text{Alloy}} = \frac{2.7 \times 900 \times 9.6}{5.4 \times 4.8} \]Calculate this to find the specific heat of the alloy:\[ c_{\text{Alloy}} = 450 \, \text{J/kg K} \]

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 91影视!

Key Concepts

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

Heat Transfer
Heat transfer is the process of energy moving from one body or substance to another, typically due to a difference in temperature. It can occur in three main ways: conduction, convection, and radiation.
  • Conduction: This happens mostly in solids where heat is transferred through direct contact of molecules. It's like a relay where molecules hit each other and pass on their energy.
  • Convection: This is primarily in fluids (liquids and gases) where the movement of the fluid itself carries heat along with it. Think of it as warm, lighter fluid rising and cold, denser fluid sinking, creating a cycle.
  • Radiation: Heat is transferred in the form of electromagnetic waves. This doesn't need any medium, so it can happen even in a vacuum.
In our exercise, the spheres are in a block of melting ice, suggesting that the primary mode of heat transfer is conduction.
Thermodynamics
Thermodynamics is the study of energy, heat, and work, and how they interrelate. It follows specific laws that govern how energy is conserved and transferred.
The exercise touches primarily on the first law of thermodynamics, which is the conservation of energy principle. It states that energy cannot be created or destroyed but only transferred or changed from one form to another.
In our exercise, the spheres are losing heat to the surrounding ice. The first law helps us understand that the lost heat from the spheres is gained by the ice, helping it melt without any net loss or gain in total energy.
Specific Gravity
Specific gravity is a measure of density and is defined as the ratio of the density of a substance to the density of a reference substance, typically water for liquids and solids. It is a dimensionless quantity.
  • Specific gravity helps describe how heavy a material is compared to water.
  • In our exercise, the specific gravity of aluminum and the alloy helps determine their masses when volumes are equal.
For instance, a specific gravity of 2.7 for aluminum implies it is 2.7 times denser than water. This information helps us calculate the mass of each sphere when combined with the volume, further aiding in heat transfer analysis.
Material Properties
Material properties define the characteristics of a substance, including thermal, mechanical, and chemical attributes. In this exercise, we're primarily concerned with thermal properties.
  • Specific Heat Capacity: This is the amount of heat needed to increase the temperature of one kilogram of a substance by one degree Celsius. It determines how a substance responds to changes in temperature.
  • Aluminum has a specific heat capacity of 900 J/kg K, meaning it requires 900 Joules to raise the temperature of one kilogram of aluminum by one Kelvin.
  • For the unknown alloy, its specific heat capacity is what we're attempting to find using heat transfer and specific gravity data.
Understanding these properties helps us compare different materials, predict behavior in thermal environments, and solve exercises like the one given, where material differences influence energy transfer.

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

To prevent misting of the windscreen of an automobile, recirculated warm air at \(37^{\circ} \mathrm{C}\) is blown over the inner surface. The windscreen glass \((k=1.0 \mathrm{~W} / \mathrm{m} \mathrm{K})\) is 4 \(\mathrm{mm}\) thick, and the ambient temperature is \(5^{\circ} \mathrm{C}\). The outside and inside heat transfer coefficients are 70 and \(35 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively. (i) Determine the temperature of the inside surface of the glass. (ii) If the air inside the automobile is at \(20^{\circ} \mathrm{C}, 1\) atm, and \(80 \%\) relative humidity, will misting occur? (Refer to your thermodynamics text for the principles of psychrometry.)

The horizontal roof of a building is surfaced with black tar paper of emittance \(0.96 .\) On a clear, still night the air temperature is \(5^{\circ} \mathrm{C}\), and the effective temperature of the sky as a black radiation \(\sin \mathrm{k}\) is \(-60^{\circ} \mathrm{C}\). The underside of the roof is well insulated. (i) Estimate the roof surface temperature for a convective heat transfer coefficient of \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). (ii) If the wind starts blowing, giving a convective heat transfer coefficient of 20 \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\), what is the new roof temperature? (iii) Repeat the preceding calculations for aluminum roofing of emittance \(0.15 .\)

A tent is pitched on a mountain in an exposed location. The tent walls are opaque to thermal radiation. On a clear night the outside air temperature is \(-1^{\circ} \mathrm{C}\), and the effective temperature of the sky as a black radiation \(\sin k\) is \(-60^{\circ} \mathrm{C}\). The convective heat transfer coefficient between the tent and the ambient air can be taken to be 8 \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\). If the temperature of the outer surface of a sleeping bag on the tent floor is measured to be \(10^{\circ} \mathrm{C}\), estimate the heat loss from the bag in \(\mathrm{W} / \mathrm{m}^{2}\), (i) if the emittance of the tent material is \(0.7 .\) (ii) if the outer surface of the tent is aluminized to give an emittance of \(0.2\). For the sleeping bag, take an emittance of \(0.8\) and a convective heat transfer coefficient of \(4 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Assume that the ambient air circulates through the tent.

A furnace wall is to operate with inner and outer surface temperatures of \(1500 \mathrm{~K}\) and \(320 \mathrm{~K}\), respectively. Insulating bricks measuring \(20 \mathrm{~cm} \times 10 \mathrm{~cm} \times 8 \mathrm{~cm}\) are available in two kinds at the same price. Type A has a thermal conductivity of \(2.0\) \(\mathrm{W} / \mathrm{m} \mathrm{K}\) and a maximum allowable temperature of \(1600 \mathrm{~K}\). Type B has a thermal conductivity of \(1.0 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and a maximum allowable temperature of \(1000 \mathrm{~K}\). Determine how the bricks should be arranged so as not to exceed a heat flow per unit area of \(1000 \mathrm{~W} / \mathrm{m}^{2}\), and minimize the cost of the walls.

A system consists of a body in which heat is continuously generated at a rate \(\underline{Q}_{v}\), while heat is lost from the body to its surroundings by convection. Using the lumped thermal capacity model, derive the differential equation governing the temperature response of the body. If the body is at temperature \(T_{o}\) when time \(t=0\), solve the differential equation to obtain \(T(t) .\) Also determine the steady-state temperature.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Geometrical and Physical Optics

Read Explanation

Rotational Dynamics

Read Explanation

Classical Mechanics

Read Explanation

Astrophysics

Read Explanation

Energy Physics

Read Explanation
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.