/*! 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 5 Find (a) the energy absorbed as ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find (a) the energy absorbed as heat and (b) the change in entropy of a \(2.00 \mathrm{~kg}\) block of copper whose temperature is increased reversibly from \(25.0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). The specific heat of copper is \(386 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\)

Short Answer

Expert verified
The energy absorbed is 57900 J and the change in entropy is 172.49 J/K.

Step by step solution

01

Convert Temperatures to Kelvin

First, we need to convert the given temperatures from degrees Celsius to Kelvin. The conversion formula is \( T(K) = T(°C) + 273.15 \).\[ T_1 = 25.0^{\circ}C + 273.15 = 298.15 \text{ K} \] \[ T_2 = 100^{\circ}C + 273.15 = 373.15 \text{ K} \]
02

Calculate Energy Absorbed as Heat

To find the energy absorbed as heat \( Q \), use the formula \( Q = mc\Delta T \), where \( m \) is the mass, \( c \) is the specific heat, and \( \Delta T \) is the change in temperature.Given: \( m = 2.00 \text{ kg} \), \( c = 386 \text{ J/kg} \cdot \text{K} \), and \( \Delta T = 373.15 \text{ K} - 298.15 \text{ K} = 75 \text{ K} \).\[ Q = 2.00 \times 386 \times 75 = 57900 \text{ J} \]
03

Calculate Change in Entropy

For a reversible process, the change in entropy \( \Delta S \) can be calculated using \( \Delta S = \frac{Q_{rev}}{T_{avg}} \), where \( T_{avg} \) is the average temperature.The average temperature \( T_{avg} = \frac{T_1 + T_2}{2} = \frac{298.15 + 373.15}{2} = 335.65 \text{ K} \).\[ \Delta S = \frac{57900}{335.65} \approx 172.49 \text{ J/K} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Heat Transfer
Heat transfer refers to the movement of thermal energy from one body or substance to another due to a temperature difference. In our exercise, the copper block experiences heat transfer as its temperature rises from 25°C to 100°C. The flow of heat is driven by the difference in temperature between the copper and its surroundings. When the block absorbs energy, this energy is what causes the atoms within the copper to vibrate more rapidly, resulting in an increase in temperature.Understanding heat transfer allows us to calculate the energy absorbed using the formula:
  • \( Q = mc\Delta T \)
Where:
  • \( Q \) = Energy absorbed as heat (in Joules)
  • \( m \) = Mass of the substance (in kg)
  • \( c \) = Specific heat capacity (in J/kg·K)
  • \( \Delta T \) = Change in temperature (in K)
In this problem, it was calculated that 57,900 Joules of energy are absorbed when the block heats up. This calculation is vital in determining how different materials react to heat and in designing systems that manage thermal energy effectively.
Specific Heat Capacity
Specific heat capacity is a fundamental property that indicates how much heat energy is required to change the temperature of a unit mass of a substance by one degree Kelvin. In simpler terms, it tells us how easily a substance can change its temperature. Copper, like other metals, has a specific heat capacity. In our example, it's given as 386 J/kg·K. Higher specific heat means that the substance requires more energy to change its temperature. Conversely, substances with a low specific heat heat up or cool down more quickly. In engineering and material science, selecting a material with the appropriate specific heat capacity is essential for thermal management.
Entropy Change
Entropy is a measure of disorder or randomness in a system. When heat is transferred into or out of a system, it affects the system's entropy. The change in entropy for a reversible process is calculated using:
  • \(\Delta S = \frac{Q_{rev}}{T_{avg}}\)
Where:
  • \(\Delta S\) = Change in entropy (in J/K)
  • \(Q_{rev}\) = Reversible heat absorbed (in Joules)
  • \(T_{avg}\) = Average temperature during the process (in Kelvin)
In our problem, the change in entropy was calculated to be approximately 172.49 J/K. This calculation confirms that as the copper block absorbs heat, its entropy increases, reflecting a rise in disorder as the molecules move more vigorously.
Reversible Process
A reversible process is an ideal concept in thermodynamics where systems change in such a way that the system and environment can be returned to their original states without any net change. In reality, perfectly reversible processes do not occur, as losses like friction make slight changes unavoidable. However, considering reversible processes helps simplify the analysis of thermodynamic systems. When a process is reversible, the exchange of energy as heat happens in such a controlled way that the system is always in a nearly perfect state of equilibrium. This characteristic makes reversible processes crucial for theoretical models, allowing for the precise calculation of entropy changes, as seen in our copper block example. Understanding reversible processes helps us know where and why irreversibilities occur, which is key in real-world energy systems design.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the efficiency of a fossil-fuel power plant that consumes 380 metric tons of coal each hour to produce useful work at the rate of \(750 \mathrm{MW}\). The heat of combustion of coal (the heat due to burning it) is \(28 \mathrm{MJ} / \mathrm{kg}\).

A \(500 \mathrm{~W}\) Carnot engine operates between constanttemperature reservoirs at \(100^{\circ} \mathrm{C}\) and \(60.0^{\circ} \mathrm{C}\). What is the rate at which energy is (a) taken in by the engine as heat and (b) exhausted by the engine as heat?

(a) During each cycle, a Carnot engine absorbs \(750 \mathrm{~J}\) as heat from a high-temperature reservoir at \(360 \mathrm{~K}\), with the low-temperature reservoir at \(280 \mathrm{~K}\). How much work is done per cycle? (b) The engine is then made to work in reverse to function as a Carnot refrigerator between those same two reservoirs. During each cycle, how much work is required to remove \(1200 \mathrm{~J}\) as heat from the low-temperature reservoir?

The motor in a refrigerator has a power of \(200 \mathrm{~W}\). If the freezing compartment is at \(270 \mathrm{~K}\) and the outside air is at \(300 \mathrm{~K}\), and assuming the efficiency of a Carnot refrigerator, what is the maximum amount of energy that can be extracted as heat from the freezing compartment in \(10.0 \mathrm{~min} ?\)

To make ice, a freezer that is a reverse Carnot engine extracts \(42 \mathrm{~kJ}\) as heat at \(-15^{\circ} \mathrm{C}\) during each cycle, with coefficient of performance \(5.7 .\) The room temperature is \(30.3^{\circ} \mathrm{C}\). How much (a) energy per cycle is delivered as heat to the room and (b) work per cycle is required to run the freezer?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.