A heat source generates heat at a rate of 71.0 W (1 W = 1 J/s). How much entropy does this produce per hour in the surroundings at 26.2 °C? Assume the heat transfer is reversible. AS =
A heat source generates heat at a rate of 71.0 W (1 W = 1 J/s). How much entropy does this produce per hour in the surroundings at 26.2 °C? Assume the heat transfer is reversible. AS =
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Problem Statement:
A heat source generates heat at a rate of 71.0 W (1 W = 1 J/s). How much entropy does this produce per hour in the surroundings at 26.2 ˚C? Assume the heat transfer is reversible.
### Solution:
To calculate the entropy change (\(\Delta S\)), we use the following relationship for a reversible process:
\[
\Delta S = \frac{Q}{T}
\]
Where:
- \( Q \) is the heat transferred.
- \( T \) is the absolute temperature (in Kelvin).
First, let's convert the given temperature from Celsius to Kelvin.
\[
T = 26.2 + 273.15 = 299.35 \, \text{K}
\]
The power (\( P \)) is given as 71.0 W, which is equivalent to 71.0 J/s. To find the total heat (\( Q \)) transferred over an hour, we convert the time to seconds:
\[
1 \, \text{hour} = 3600 \, \text{seconds}
\]
Then calculate \( Q \):
\[
Q = P \times \text{time} = 71.0 \, \text{W} \times 3600 \, \text{s} = 255600 \, \text{J}
\]
Now, using the entropy change formula:
\[
\Delta S = \frac{255600 \, \text{J}}{299.35 \, \text{K}} \approx 853.85 \, \text{J/K}
\]
Thus, the entropy change (\(\Delta S\)) produced per hour in the surroundings is:
\[
\Delta S = \boxed{853.85} \, \text{J/K}
\]
### Diagram
There are no diagrams or graphs provided in this problem statement. Just a calculation shown in the form of text.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbd6e2cac-a13e-4af0-8d42-5fb7a66b94d3%2F5892224c-be97-481f-b19d-3f95f5cf1513%2Fo4n3lre_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
A heat source generates heat at a rate of 71.0 W (1 W = 1 J/s). How much entropy does this produce per hour in the surroundings at 26.2 ˚C? Assume the heat transfer is reversible.
### Solution:
To calculate the entropy change (\(\Delta S\)), we use the following relationship for a reversible process:
\[
\Delta S = \frac{Q}{T}
\]
Where:
- \( Q \) is the heat transferred.
- \( T \) is the absolute temperature (in Kelvin).
First, let's convert the given temperature from Celsius to Kelvin.
\[
T = 26.2 + 273.15 = 299.35 \, \text{K}
\]
The power (\( P \)) is given as 71.0 W, which is equivalent to 71.0 J/s. To find the total heat (\( Q \)) transferred over an hour, we convert the time to seconds:
\[
1 \, \text{hour} = 3600 \, \text{seconds}
\]
Then calculate \( Q \):
\[
Q = P \times \text{time} = 71.0 \, \text{W} \times 3600 \, \text{s} = 255600 \, \text{J}
\]
Now, using the entropy change formula:
\[
\Delta S = \frac{255600 \, \text{J}}{299.35 \, \text{K}} \approx 853.85 \, \text{J/K}
\]
Thus, the entropy change (\(\Delta S\)) produced per hour in the surroundings is:
\[
\Delta S = \boxed{853.85} \, \text{J/K}
\]
### Diagram
There are no diagrams or graphs provided in this problem statement. Just a calculation shown in the form of text.
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