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Chapter 14: Problem 23

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 \( 239.6 \mathrm{~J/(kg \cdot C^{\circ})} \).

Step by step solution

01

Understanding the Given Values

We are given the following: the weight of the sample, \( W = 28.4 \mathrm{~N} \), the heat added, \( Q = 1.25 \times 10^4 \mathrm{~J} \), and the change in temperature, \( \Delta T = 18.0 \mathrm{~C}^{\circ} \). We need to find the specific heat \( c \) of the sample.
02

Calculating the Mass

First, convert weight (force) to mass using the formula \( W = mg \) where \( g = 9.8 \mathrm{~m/s^2} \). Rearrange to find \( m = \frac{W}{g} = \frac{28.4 \mathrm{~N}}{9.8 \mathrm{~m/s^2}} \approx 2.9 \mathrm{~kg} \).
03

Using the Specific Heat Formula

The formula for specific heat is \( c = \frac{Q}{m \Delta T} \). Substitute the respective values: \( Q = 1.25 \times 10^4 \mathrm{~J} \), \( m = 2.9 \mathrm{~kg} \), and \( \Delta T = 18.0 \mathrm{~C}^{\circ} \) into the equation.
04

Calculating Specific Heat

Put the values into the formula: \[ c = \frac{1.25 \times 10^4 \mathrm{~J}}{2.9 \mathrm{~kg} \times 18.0 \mathrm{~C}^{\circ}} \approx 239.6 \mathrm{~J/(kg \cdot C^{\circ})} \].
05

Verifying Units

Check that the final units of specific heat are \( \mathrm{J/(kg \cdot C^{\circ})} \), which are the correct and standard units for specific heat capacity in the International System of Units (SI).

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Key Concepts

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

Specific Heat
Specific heat is a fundamental concept in physics that describes how much energy is needed to change the temperature of a unit mass of a substance by one degree. It's symbolized as \(c\) and expressed in the units \(\mathrm{J/(kg \cdot C^{\circ})}\).
Specific heat varies depending on the material. This means different materials require different amounts of energy to change their temperatures.
  • Higher specific heat means more energy is needed for temperature change.
  • Lower specific heat indicates less energy needed.
This property is crucial in understanding how substances heat up and cool down in our everyday environment.
Heat Energy
Heat energy, often represented as \(Q\), is the energy transferred, typically as a result of temperature differences. In physics, it is measured in joules (\(\mathrm{J}\)).
  • Heat can be transferred by conduction, convection, and radiation.
  • It's a form of energy that can cause changes in the temperature of substances.
During our exercise, the metal sample absorbed \(1.25 \times 10^4\) joules of heat. Understanding how this heat energy affects the metal's temperature is key to solving the problem of specific heat.
Temperature Change
Temperature change, denoted as \(\Delta T\), is the difference in the initial and final temperatures of a substance after heat has been applied. It is measured in degrees Celsius (\(\mathrm{C^{\circ}}\)).
This concept is a direct indicator of the effect of heat energy on a substance.
  • A small temperature change might suggest a high specific heat.
  • A large temperature change often indicates a lower specific heat.
In our metal sample exercise, the temperature increased by \(18.0 \mathrm{C^{\circ}}\). Calculating this change helps us better estimate the specific heat when combined with other measurements.
Physics Problem Solving
Physics problem solving requires a structured approach to understand and resolve physical phenomena and their mathematical representations. In our exercise, this involves several key steps:
  • Identifying given values (weight, heat energy, temperature change).
  • Converting weight to mass, using the relationship between force and mass \(W = mg\).
  • Applying the specific heat formula \(c = \frac{Q}{m \Delta T}\).
  • Substituting known values into the formula to compute the result.
This systematic approach ensures accuracy and understanding of the physical concepts involved. By organizing and calculating each component step-by-step, like we did, you can effectively solve similar physics problems in the future.

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Most popular questions from this chapter

A \(5.00 \mathrm{~kg}\) lead sphere is dropped from the top of a \(60.0-\mathrm{m}\) -tall building. If all of its kinetic energy is converted into heat when it hits the sidewalk, how much will its temperature rise? (Ignore air resistance.)

(a) At what temperature do the Fahrenheit and Celsius scales give the same reading? (b) Is there any temperature at which the Kelvin and Celsius scales coincide?

(a) A typical student listening attentively to a physics lecture has a heat output of \(100 \mathrm{~W}\). How much heat energy does a class of 90 physics students release into a lecture hall over the course of a 50 min lecture? (b) Assume that all the heat energy in part (a) is transferred to the \(3200 \mathrm{~m}^{3}\) of air in the room. The air has a specific heat of \(1020 \mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K})\) and a density of \(1.20 \mathrm{~kg} / \mathrm{m}^{3} .\) If none of the heat escapes and the air- conditioning system is off, how much will the temperature of the air in the room rise during the 50 min lecture? (c) If the class is taking an exam, the heat output per student rises to \(280 \mathrm{~W}\). What is the temperature rise during \(50 \mathrm{~min}\) in this case?

You have no doubt noticed that you usually shiver when you get out of the shower. Shivering is the body's way of generating heat to restore its internal temperature to the normal \(37^{\circ} \mathrm{C}\). and it produces approximately \(290 \mathrm{~W}\) of heat power per square meter of body area. A \(68 \mathrm{~kg}(150 \mathrm{lb}), 1.78 \mathrm{~m}(5 \mathrm{ft}, 10\) in.) person has approximately \(1.8 \mathrm{~m}^{2}\) of surface area. How long would this person have to shiver to raise his or her body temperature by \(1.0 \mathrm{C}^{\circ},\) assuming that none of this heat is lost by the body? The specific heat of the body is about \(3500 \mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K}).\)

A spherical pot of hot coffee contains \(0.75 \mathrm{~L}\) of liquid (essentially water) at an initial temperature of \(95^{\circ} \mathrm{C}\). The pot has an emissivity of \(0.60,\) and the surroundings are at a temperature of \(20.0^{\circ} \mathrm{C}\). Calculate the coffee's rate of heat loss by radiation.
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