/*! 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 99 A block of material has a mass o... [FREE SOLUTION] | 91Ó°ÊÓ

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A block of material has a mass of \(130 \mathrm{kg}\) and a volume of \(4.6 \times 10^{-2} \mathrm{m}^{3} .\) The material has a specific heat capacity and coefficient of volume expansion, respectively, of \(750 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and \(6.4 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1}\) How much heat must be added to the block in order to increase its volume by \(1.2 \times 10^{-5} \mathrm{m}^{3} ?\)

Short Answer

Expert verified
The heat required is calculated using the formulas for volumetric expansion and heat transfer.

Step by step solution

01

Understand the Problem

The problem involves calculating the amount of heat needed to cause a specific increase in volume of a material using its coefficient of volume expansion. We know the mass, volume, specific heat capacity, and coefficient of volume expansion of the material.
02

Establish a Relationship Between Volume Change and Temperature Change

Use the formula for volumetric expansion: \[ \Delta V = \beta V_0 \Delta T \],where \( \Delta V \) is the change in volume, \( \beta \) is the coefficient of volume expansion, \( V_0 \) is the initial volume, and \( \Delta T \) is the change in temperature.
03

Calculate the Temperature Change Required

Rearrange the volumetric expansion formula to find \( \Delta T \): \[ \Delta T = \frac{\Delta V}{\beta V_0} \] Plug in the given values: \[ \Delta T = \frac{1.2 \times 10^{-5} \ m^3}{6.4 \times 10^{-5} \ ^{\circ}C^{-1} \cdot 4.6 \times 10^{-2} \ m^3} \]. Calculate \( \Delta T \).
04

Calculate the Heat Required Using Specific Heat Formula

Use the formula for heat transfer: \[ Q = mc\Delta T \], where \( Q \) is the heat added, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change calculated earlier. Insert the given values and compute \( Q \).
05

Final Calculation and Solution

Substitute all known values:\[ m = 130 \, \text{kg}, \quad c = 750 \, \text{J/kg}\cdot^{\circ}C, \quad \Delta T \text{from Step 3}.\]Calculate \( Q \) using the heat transfer formula.

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Key Concepts

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

Understanding Specific Heat Capacity
Specific heat capacity is a property that tells us how much heat energy needs to be added to a substance to change its temperature by one degree Celsius per unit mass.
It's measured in joules per kilogram per degree Celsius \(J/(kg \cdot °C) \). When you know the specific heat capacity of a material, you can predict how much its temperature will change when a certain amount of heat is added.
Here, for this material block, the specific heat capacity is given as 750 J/(kg°C).
In simpler terms, a higher specific heat capacity means the material can absorb more heat without a large change in temperature. This concept is crucial in calculating the amount of heat required to change the volume of a block because every bit of heat we provide changes the temperature, which in turn affects the volume due to thermal expansion.
Understanding this concept helps relate heat input to temperature changes in any thermal dynamics problem.
Explaining Coefficient of Volume Expansion
The coefficient of volume expansion is a measure of how much the volume of a material changes with temperature.
It is expressed as \(\beta\), and for our material, \(\beta = 6.4 \times 10^{-5} \ \text{°C}^{-1}\).
This means for every increase of one degree Celsius in the material's temperature, its volume will increase by a factor of \(6.4 \times 10^{-5}\) times its original volume.
This property is key when dealing with volume changes due to heat. Furthermore, it directly relates to how much temperature change is needed to achieve a certain change in volume.
Using the combined knowledge of specific heat capacity and the coefficient of volume expansion allows us to determine how much heat leads to a specific volume increase. The formula \(\Delta V = \beta V_0 \Delta T\) connects these concepts by relating the temperature change that causes a change in volume.
Essentials of Heat Transfer
Heat transfer is the movement of heat energy from one place or material to another.
In our scenario, heat transfer is responsible for increasing the temperature of the block, which consequently causes it to expand.
The heat energy required to cause this volume expansion can be calculated using the formula \(Q = mc\Delta T\). Here, \(Q\) is the total heat energy, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change.
These calculations rely on understanding that heat transfer into the block results in an increase in the internal energy, manifested through temperature rise and expansion.
For practical purposes:
  • Determine the energy needed to raise the temperature first.
  • Apply this to the formula to calculate the heat needed.
  • This approach keeps calculations organized and easy to understand.
With these principles, students can better grasp how heat energy impacts physical changes in volume and temperature.

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Most popular questions from this chapter

The column of mercury in a barometer (see Figure 11.11 ) has a height of \(0.760 \mathrm{m}\) when the pressure is one atmosphere and the temperature is \(0.0^{\circ} \mathrm{C}\). Ignoring any change in the glass containing the mercury, what will be the height of the mercury column for the same one atmosphere of pressure when the temperature rises to \(38.0^{\circ} \mathrm{C}\) on a hot day?

A person eats a container of strawberry yogurt. The Nutritional Facts label states that it contains 240 Calories \((1\) Calorie \(=4186 \mathrm{J})\). What mass of perspiration would one have to lose to get rid of this energy? At body temperature, the latent heat of vaporization of water is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}\).

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?

The Eiffel Tower is a steel structure whose height increases by \(19.4 \mathrm{cm}\) when the temperature changes from -9 to \(+41^{\circ} \mathrm{C}\) . What is the approximate height (in meters) at the lower temperature?

A steel bicycle wheel (without the rubber tire) is rotating freely with an angular speed of \(18.00 \mathrm{rad} / \mathrm{s}\). The temperature of the wheel changes from \(-100.0\) to \(+300.0^{\circ} \mathrm{C} .\) No net external torque acts on the wheel, and the mass of the spokes is negligible. (a) Does the angular speed increase or decrease as the wheel heats up? Why? (b) What is the angular speed at the higher temperature?

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