/*! 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 is approximately 239.46 J/(kg°C).

Step by step solution

01

Understand the Units and Concepts

First, recognize that you need to calculate specific heat, which uses the formula: \( q = mc\Delta T \). Here, \( q \) is the heat added (in Joules), \( m \) is the mass (in kg), \( c \) is the specific heat (in \( J/(kg \cdot °C) \)), and \( \Delta T \) is the temperature change (in °C).
02

Convert Weight to Mass

Since the weight of the sample is given as 28.4 N, convert this weight into mass using the equation \( W = mg \), where \( g \) (acceleration due to gravity) is 9.8 m/s². Rearrange to find mass: \( m = \frac{W}{g} = \frac{28.4}{9.8} \approx 2.9 \) kg.
03

Apply the Specific Heat Formula

Now that you have all the necessary values, plug them into the specific heat formula: \( q = mc\Delta T \). Rearrange this to solve for specific heat \( c \): \( c = \frac{q}{m \cdot \Delta T} = \frac{1.25 \times 10^{4} \text{ J}}{2.9 \text{ kg} \cdot 18 °C} \).
04

Calculate the Specific Heat

Perform the calculation: \( c = \frac{1.25 \times 10^{4}}{2.9 \times 18} \approx \frac{12500}{52.2} \approx 239.46 \text{ J/(kg} \cdot °C) \).

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Key Concepts

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

Heat Energy
In everyday life, heat energy is the energy transferred from one object to another because of a temperature difference. It's a key concept in understanding how substances change temperature when energy is added or removed. Heat energy in physics is usually measured in Joules (J), as in our exercise where the metal sample receives 12,500 J.

When heat energy is added to a substance, it can cause the substance's particles to move faster, which is often experienced as a temperature increase. This is precisely what happens in our problem. By understanding how heat energy is transferred and measured, students can better grasp the relationship between energy input and temperature changes.
  • Heat energy is crucial for changing the temperature of a substance.
  • Measured in Joules, it's a quantitative way to understand energy exchanges.
  • Understanding heat energy helps in solving problems where temperatures and states change.
Mass Conversion
Mass conversion is essential when working with physical quantities in physics problems. When the weight of an object is given, it's typically measured in Newtons (N). However, to calculate specific heat, we need the mass in kilograms (kg). This conversion is done by dividing the weight by the acceleration due to gravity, which is approximately 9.8 m/s² on Earth.

In the exercise, the weight of the metal sample is 28.4 N. By converting this to mass, we find:\[m = \frac{W}{g} = \frac{28.4}{9.8} \approx 2.9 \text{ kg}\]Using the correct mass is vital for accurately computing the specific heat or any other property that involves mass.
  • Recognize weight as force and convert it to mass.
  • Use Earth's gravitational acceleration (9.8 m/s²) for conversion.
  • Ensure mass is in kg for accurate calculations in physics.
Temperature Change
Temperature change (\(\Delta T\)) is another critical component in these calculations. It represents the difference in temperature before and after heat energy is added. In our problem, the metal sample's temperature rose by 18°C after adding heat. This change is used in the specific heat formula to determine how much heat energy is required to change a substance's temperature.

Temperature change helps us understand how different materials respond to energy transfer. Some materials heat up quickly, while others require more energy to change their temperature, indicating a higher specific heat.
  • Temperature change is the difference in initial and final temperatures.
  • It helps determine the energy needed for temperature change.
  • Key in finding specific heat capacity.
Physics Problem Solving
Physics problem solving involves understanding and applying various concepts and formulas to find a solution. In our exercise, the task is to find the specific heat of a metal sample using given values. This involves several steps, including unit conversion, understanding formulas, and performing calculations.

Here's a breakdown of the process:
  • Identify and understand the formula: Specific heat, \( q = mc\Delta T \)
  • Convert given units when necessary, such as weight to mass.
  • Plug known values into the formula and rearrange to find the unknown.
  • Perform arithmetic calculations accurately.
  • Draw conclusions from the computed results.
By breaking down the process into clear steps and consistently applying known formulas and concepts, solving physics problems becomes more manageable and systematic.

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Most popular questions from this chapter

A Walk in the Sun. Consider a poor lost soul walking at 5 \(\mathrm{km} / \mathrm{h}\) on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of \(280 \mathrm{W},\) and almost all of this energy is con- verted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to \(k^{\prime} A_{\text { skin }}\left(T_{\text { air }}-T_{\text { skin }}\right),\) where \(k^{\prime}\) is \(54 \mathrm{J} / \mathrm{h} \cdot \mathrm{C}^{\circ} \cdot \mathrm{m}^{2},\) the exposed skin area \(A_{\text { skin }}\) is \(1.5 \mathrm{m}^{2},\) the air temperature \(T_{\mathrm{air}}\) is \(47^{\circ} \mathrm{C},\) and the skin temperature \(T_{\text { skin }}\) is \(36^{\circ} \mathrm{C} ;\) (iii) the skin absorbs radiant energy from the sun at a rate of 1400 \(\mathrm{W} / \mathrm{m}^{2}\) ; (iv) the skin absorbs radiant energy from the environment, which has temperature \(47^{\circ} \mathrm{C}\) . (a) Calculate the net rate (in watts) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is \(e=1\) and that the skin temperature is initially \(36^{\circ} \mathrm{C}\) . Which mechanism is the most important? (b) At what rate (in \(\mathrm{L} / \mathrm{h} )\) must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at \(36^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\)) (c) Suppose instead the person is protected by light-colored clothing \((e \approx 0)\) so that the exposed skin area is only 0.45 \(\mathrm{m}^{2} .\) What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.

A nail driven into a board increases in temperature. If we assume that 60\(\%\) of the kinetic energy delivered by a 1.80-kg hammer with a speed of 7.80 \(\mathrm{m} / \mathrm{s}\) is transformed into heat that flows into the nail and does not flow out, what is the temperature increase of an \(8.00-\mathrm{g}\) aluminum nail after it is struck ten times?

A copper calorimeter can with mass 0.446 \(\mathrm{kg}\) contains 0.0950 \(\mathrm{kg}\) of ice. The system is initially at \(0.0^{\circ} \mathrm{C} .\) (a) If 0.0350 \(\mathrm{kg}\) of steam at \(100.0^{\circ} \mathrm{C}\) and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents? (b) At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

You pour 108 \(\mathrm{cm}^{3}\) of ethanol, at a temperature of \(-10.0^{\circ} \mathrm{C},\) into a graduated cylinder initially at \(20.0^{\circ} \mathrm{C},\) filling it to the very top. The cylinder is made of glass with a specific heat of 840 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and a coefficient of volume expansion of \(1.2 \times 10^{-5} \mathrm{K}^{-1} ;\) its mass is 0.110 \(\mathrm{kg} .\) The mass of the ethanol is 0.0873 \(\mathrm{kg} .\) (a) What will be the final temperature of the ethanol, once thermal equilibrium is reached? (b) How much ethanol will overflow the cylinder before thermal equilibrium is reached?

A crate of fruit with mass 35.0 \(\mathrm{kg}\) and specific heat 3650 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) slides down a ramp inclined at \(36.9^{\circ}\) below the horizontal. The ramp is 8.00 \(\mathrm{m}\) long. (a) If the crate was at rest at the top of the incline and has a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) at the bottom, how much work was done on the crate by friction? (b) If an amount of heat equal to the magnitude of the work done by friction goes into the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change?

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