/*! 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 You are given a sample of metal ... [FREE SOLUTION] | 91Ó°ÊÓ

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You are given a sample of metal and asked to determine its specific heat. You weigh the sample and find that its weight is \(28.4 \mathrm{~N}\). You carefully add \(1.25 \times 10^{4} \mathrm{~J}\) of heat energy to the sample and find that its temperature rises \(18.0 \mathrm{C}^{\circ} .\) What is the sample's specific heat?

Short Answer

Expert verified
The specific heat of the sample is approximately \( 235 \, \mathrm{J/kg \cdot C} \)

Step by step solution

01

Convert weight into mass

First the weight given in Newton needs to be converted into kilograms (kg) using the formula \( m = \frac{w}{g} \) where \( m \) is the mass, \( w \) is the weight in Newtons (N), and \( g \) is the acceleration due to gravity which is approximately \( 9.8 \, \mathrm{ms}^{-2} \). Therefore, the mass of the metal in kilograms is calculated as follows: \( m = \frac{28.4 \, \mathrm{N}}{9.8 \, \mathrm{ms}^{-2}} \approx 2.90 \, \mathrm{kg} \)
02

Calculate heat energy

The heat energy is given by the problem and we simply use the provided energy: \( Q = 1.25 \times 10^{4} \, \mathrm{J} \) (Joules).
03

Calculate the change in temperature

The temperature change is also given by the problem as \( \Delta T = 18.0 \, \mathrm{C} \)
04

Solve for specific heat capacity

To calculate the specific heat, use the formula \( c = \frac{Q}{m \Delta T} \). Substituting the known values into this equation gives: \( c = \frac{1.25 \times 10^{4} \, \mathrm{J}}{2.90 \, \mathrm{kg} \times 18.0 \, \mathrm{C}} \approx 235 \, \mathrm{J/kg \cdot C} \)

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Key Concepts

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

Heat Energy
Heat energy is the energy that is transferred from one system to another due to a difference in temperature. In this exercise, heat energy flows into the metal sample, causing its temperature to rise. We measure this energy in Joules (J).
To calculate the change in temperature due to the addition of heat, it's important to know the amount of heat energy provided and how it interacts with the material. Heat energy accounts for the total energy transferred during processes like heating, melting, or evaporating.
Understanding heat energy helps us determine how substances change temperature and phase, which is fundamental in fields like physics, chemistry, and engineering.
  • Measurement Unit: Joules (J)
  • Role: Increases temperature or changes state.
Mass Conversion
Mass conversion is an essential step in measuring the specific heat of a substance when its weight is given in Newtons. Weight is the force exerted by gravity on an object, and it varies according to the gravitational pull of a given environment.
To determine mass from weight, we use the formula: \[ m = \frac{w}{g} \] where \( w \) is the weight in Newtons and \( g \) is the acceleration due to gravity, roughly \( 9.8 \, \mathrm{m/s}^2 \) on Earth. This calculation gives us mass in kilograms, which is necessary for calculating specific heat.
  • Formula: \( m = \frac{w}{g} \)
  • Weight to Mass: Ensures consistent units for calculations.
Mass conversion is crucial because it transforms weight into mass, allowing us to consistently use mass with other SI units in scientific equations, such as those for heat energy.
Temperature Change
Temperature change (\(\Delta T\)) is the difference between the initial and final temperature of a substance after heat energy has been added or removed. It is typically measured in degrees Celsius (°C) or Kelvin (K), depending on the problem context.
In the exercise, we find that the temperature of the metal sample rises by \(18.0 \, \mathrm{C}^{\circ}\). This temperature change provides information about how the sample absorbs energy.
Understanding temperature change is vital because it tells us about the material's thermal response to energy fluctuations. More specifically:
  • High temperature change: Indicates significant energy absorption.
  • Low temperature change: Suggests less energy absorption or higher resistance to temperature change.
These insights are powerful in thermodynamics, where temperature change can indicate various material properties and behaviors.
Specific Heat Capacity Formula
Specific heat capacity (\(c\)) is a crucial property that describes how much energy a substance needs to change its temperature. To find it, we use the formula: \[ c = \frac{Q}{m \Delta T} \]
Where:
  • \(Q\): Heat energy added (Joules)
  • \(m\): Mass of the substance (kilograms)
  • \(\Delta T\): Temperature change (Celsius or Kelvin)
This formula helps us calculate the specific heat, measured in \(J/\mathrm{kg} \cdot \mathrm{C}\). It tells us how much energy is needed per kilogram to raise the temperature by one degree Celsius.
In the exercise provided, substituting known values results in a specific heat of approximately \(235 \, \mathrm{J/kg \cdot C}\). Specific heat capacity is critically informative for practical applications such as material selection in engineering, where knowing how materials respond to heat influences design and safety.

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Most popular questions from this chapter

A copper sphere with density \(8900 \mathrm{~kg} / \mathrm{m}^{3},\) radius \(5.00 \mathrm{~cm}\) and emissivity \(e=1.00\) sits on an insulated stand. The initial temperature of the sphere is \(300 \mathrm{~K}\). The surroundings are very cold, so the rate of absorption of heat by the sphere can be neglected. (a) How long does it take the sphere to cool by \(1.00 \mathrm{~K}\) due to its radiation of heat energy? Neglect the change in heat current as the temperature decreases. (b) To assess the accuracy of the approximation used in part (a), what is the fractional change in the heat current \(H\) when the temperature changes from \(300 \mathrm{~K}\) to \(299 \mathrm{~K} ?\)

A glass flask whose volume is \(1000.00 \mathrm{~cm}^{3}\) at \(0.0^{\circ} \mathrm{C}\) is completely filled with mercury at this temperature. When flask and mercury are warmed to \(55.0^{\circ} \mathrm{C}, 8.95 \mathrm{~cm}^{3}\) of mercury overflow. If the coefficient of volume expansion of mercury is \(18.0 \times 10^{-5} \mathrm{~K}^{-1},\) compute the coefficient of volume expansion of the glass.

To measure the specific heat in the liquid phase of a newly developed cryoprotectant, you place a sample of the new cryoprotectant in contact with a cold plate until the solution's temperature drops from room temperature to its freezing point. Then you measure the heat transferred to the cold plate. If the system isn't sufficiently isolated from its roomtemperature surroundings, what will be the effect on the measurement of the specific heat? (a) The measured specific heat will be greater than the actual specific heat; (b) the measured specific heat will be less than the actual specific heat; (c) there will be no effect because the thermal conductivity of the cryoprotectant is so low; (d) there will be no effect on the specific heat, but the temperature of the freezing point will change.

One end of an insulated metal rod is maintained at \(100.0^{\circ} \mathrm{C}\), and the other end is maintained at \(0.00^{\circ} \mathrm{C}\) by an ice-water mixture. The rod is \(60.0 \mathrm{~cm}\) long and has a cross-sectional area of \(1.25 \mathrm{~cm}^{2}\). The heat conducted by the rod melts \(8.50 \mathrm{~g}\) of ice in \(10.0 \mathrm{~min} .\) Find the thermal conductivity \(k\) of the metal.

A \(25,000 \mathrm{~kg}\) subway train initially traveling at \(15.5 \mathrm{~m} / \mathrm{s}\) slows to a stop in a station and then stays there long enough for its brakes to cool. The station's dimensions are \(65.0 \mathrm{~m}\) long by \(20.0 \mathrm{~m}\) wide by \(12.0 \mathrm{~m}\) high. Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, by how much does the air temperature in the station rise? Take the density of the air to be \(1.20 \mathrm{~kg} / \mathrm{m}^{3}\) and its specific heat to be \(1020 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

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