/*! 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 27 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 N. You carefully add \(1.25 \times 10{^4}\) J of heat energy to the sample and find that its temperature rises 18.0 C\(^\circ\). What is the sample's specific heat?

Short Answer

Expert verified
The specific heat of the sample is approximately 240.45 J/kg·°C.

Step by step solution

01

Understanding Given Data

First, identify the given data from the problem. We have the weight of the sample as 28.4 N, the heat energy added as \(1.25 \times 10^4\) J, and the temperature change as 18.0 C\(^\circ\).
02

Relate Weight to Mass

We need the mass of the metal sample, which can be found using its weight. Recall that weight (\(W\)) is related to mass (\(m\)) by the equation: \(W = m \cdot g\), where \(g = 9.81 \text{ m/s}^2\). Rearrange to find \(m = \frac{W}{g}\). Substitute in: \(m = \frac{28.4}{9.81}\).
03

Calculate Mass

Perform the calculation from Step 2 to find the mass of the sample:\[ m = \frac{28.4}{9.81} \approx 2.896 \text{ kg} \].
04

Use the Specific Heat Formula

The formula for specific heat \(c\) is \(c = \frac{Q}{m \cdot \Delta T}\), where \(Q\) is the heat added, \(m\) is the mass, and \(\Delta T\) is the temperature change. We know \(Q = 1.25 \times 10^4 \text{ J}\), \(m\) is approximately 2.896 kg, and \(\Delta T = 18.0 \text{ C}^\circ\).
05

Calculate Specific Heat

Substitute the known values into the specific heat formula from Step 4:\[ c = \frac{1.25 \times 10^4}{2.896 \times 18.0} \approx 240.45 \text{ J/kg}\cdot\text{C}^\circ \].

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Key Concepts

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

Specific Heat Capacity
The concept of specific heat capacity is essential in understanding how substances absorb and transfer heat. It is defined as the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius. In formula terms, it is represented as:
  • \[ c = \frac{Q}{m \cdot \Delta T} \]
Here, \( c \) is the specific heat capacity, \( Q \) is the heat energy added, \( m \) is the mass of the substance, and \( \Delta T \) is the temperature change.
In essence, specific heat capacity is a measure of a material's ability to store thermal energy. Materials with high specific heat capacities can absorb a lot of heat without a significant change in temperature. This property is crucial in applications such as cooking and climate regulation.
Knowing a material's specific heat capacity helps in designing systems that need efficient thermal management and conservation of energy.
Heat Energy
Heat energy, often denoted as \( Q \), is the energy transferred to a substance to cause a temperature change. It is measured in Joules (J) and is a fundamental concept in thermodynamics. In our specific heat equation, \( Q \) plays a critical role as it represents the energy input to change the temperature of the given mass.
The relationship between heat energy and specific heat is key to calculating how energy is stored or released within a substance. In practical terms, when you add heat energy to a material, if the material has a high specific heat, the temperature change \( \Delta T \) will be small for the amount of heat added.
Understanding how heat energy affects temperature and phase changes is crucial in both everyday applications and scientific experiments.
Temperature Change
Temperature change, denoted \( \Delta T \), is simply the difference between the final temperature and the initial temperature of the substance. In this exercise, the temperature change was noted to be 18.0 C\(^\circ\). This value is vital because it directly influences how much heat energy a material requires to reach a certain temperature.
As seen in the specific heat formula, \( \Delta T \) is proportional to the heat energy added, provided that the mass and specific heat capacity remain constant. Thus, when we know \( \Delta T \), along with the mass and specific heat capacity, we can easily calculate how much energy was required for the temperature change.
Temperature change also helps assess the efficiency of thermal systems, whether for heating buildings, cooling engines, or conducting scientific research.
Mass Calculation
The calculation of mass from a known weight is critical in thermodynamic calculations. Weight is the force exerted by gravity on an object, and it is calculated using the formula:
  • \[ W = m \cdot g \]
where \( W \) is the weight, \( m \) is the mass, and \( g \) is the acceleration due to gravity (approximately 9.81 m/s\(^2\) on Earth).
To find the mass from weight, you rearrange the formula to solve for \( m \):
  • \[ m = \frac{W}{g} \]
In the original exercise, the specific mass of the metal was identified to be about 2.896 kg. Correctly calculating the mass is essential because it directly impacts the accuracy of the specific heat calculation.
In scientific experiments, precision in mass calculation ensures that all subsequent calculations involving heat transfer and temperature changes are as accurate as possible.

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Most popular questions from this chapter

A 4.00-kg silver ingot is taken from a furnace, where its temperature is 750.0\(^\circ\)C, and placed on a large block of ice at 0.0\(^\circ\)C. Assuming that all the heat given up by the silver is used to melt the ice, how much ice is melted?

In a container of negligible mass, 0.200 kg of ice at an initial temperature of -40.0\(^\circ\)C is mixed with a mass m of water that has an initial temperature of 80.0\(^\circ\)C. No heat is lost to the surroundings. If the final temperature of the system is 28.0\(\circ\)C, what is the mass m of the water that was initially at 80.0\(^\circ\)C?

A copper calorimeter can with mass 0.446 kg contains 0.0950 kg of ice. The system is initially at 0.0\(^\circ\)C. (a) If 0.0350 kg of steam at 100.0\(^\circ\)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?

CP A 15.0-g bullet traveling horizontally at 865 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m/s. What is the maximum temperature increase that the water could have as a result of this event?

Convert the following Celsius temperatures to Fahrenheit: (a) -62.8\(^\circ\)C, the lowest temperature ever recorded in North America (February 3, 1947, Snag, Yukon); (b) 56.7\(^\circ\)C, the highest temperature ever recorded in the United States (July 10, 1913, Death Valley, California); (c) 31.1\(^\circ\)C, the world’s highest average annual temperature (Lugh Ferrandi, Somalia).

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