### 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:\[\Delta S = \boxed{853.85} \, \text{J/K}\]### DiagramThere are no diagrams or graphs provided in this problem statement. Just a calculation shown in the form of text.

The value of entropy change is as follows.

Answered: A heat source generates heat at a rate…

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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Question

### 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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