Calculate AS when two iron blocks, each of mass 1.00 kg, one at 200 K and the other at 25 K, are placed in contact in an isolated container. The specific heat capacity of iron is 0.449 J K-¹g-¹ and may be assumed constant over the temperature range involved.
Calculate AS when two iron blocks, each of mass 1.00 kg, one at 200 K and the other at 25 K, are placed in contact in an isolated container. The specific heat capacity of iron is 0.449 J K-¹g-¹ and may be assumed constant over the temperature range involved.
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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![**Problem Statement: Entropy Change Calculation**
Calculate \( \Delta S \) when two iron blocks, each of mass 1.00 kg, one at 200 K and the other at 25 K, are placed in contact in an isolated container. The specific heat capacity of iron is 0.449 J K\(^{-1}\) g\(^{-1}\) and may be assumed constant over the temperature range involved.
---
To solve this problem, consider that both blocks will eventually reach thermal equilibrium. The process involves calculating the entropy change (\( \Delta S \)) for each block as they reach the final equilibrium temperature, considering the conservation of energy, and adding these changes to find the total entropy change.
**Key Concepts:**
1. **Isolated System:** No heat is exchanged with the surroundings.
2. **Equation for Entropy Change (\( \Delta S \))** for a block:
\[
\Delta S = m \cdot c \cdot \ln \left(\frac{T_f}{T_i}\right)
\]
where \( m \) is mass, \( c \) is specific heat, \( T_f \) is the final temperature, and \( T_i \) is the initial temperature.
3. **Equilibrium Temperature Calculation:**
Using energy balance, calculate the final temperature \( T_f \) by equating heat lost by the hot block to heat gained by the cold block:
\[
m \cdot c \cdot (T_{f} - 200) = m \cdot c \cdot (T_{f} - 25)
\]
4. **Summation of Entropy Changes:**
Total \( \Delta S_{\text{total}} = \Delta S_{\text{hot block}} + \Delta S_{\text{cold block}} \).
Use these equations and principles to determine \( \Delta S \) for the system.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d6b1f03-b4bd-4f60-ae23-d8256607755b%2F5b2f2d9e-8b34-4838-91f6-c9d5bbdb2546%2Fb276dwt_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement: Entropy Change Calculation**
Calculate \( \Delta S \) when two iron blocks, each of mass 1.00 kg, one at 200 K and the other at 25 K, are placed in contact in an isolated container. The specific heat capacity of iron is 0.449 J K\(^{-1}\) g\(^{-1}\) and may be assumed constant over the temperature range involved.
---
To solve this problem, consider that both blocks will eventually reach thermal equilibrium. The process involves calculating the entropy change (\( \Delta S \)) for each block as they reach the final equilibrium temperature, considering the conservation of energy, and adding these changes to find the total entropy change.
**Key Concepts:**
1. **Isolated System:** No heat is exchanged with the surroundings.
2. **Equation for Entropy Change (\( \Delta S \))** for a block:
\[
\Delta S = m \cdot c \cdot \ln \left(\frac{T_f}{T_i}\right)
\]
where \( m \) is mass, \( c \) is specific heat, \( T_f \) is the final temperature, and \( T_i \) is the initial temperature.
3. **Equilibrium Temperature Calculation:**
Using energy balance, calculate the final temperature \( T_f \) by equating heat lost by the hot block to heat gained by the cold block:
\[
m \cdot c \cdot (T_{f} - 200) = m \cdot c \cdot (T_{f} - 25)
\]
4. **Summation of Entropy Changes:**
Total \( \Delta S_{\text{total}} = \Delta S_{\text{hot block}} + \Delta S_{\text{cold block}} \).
Use these equations and principles to determine \( \Delta S \) for the system.
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