College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Question:**
A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both the length and the diameter of this bar are doubled, the rate at which it will conduct heat between these reservoirs will be:
- 12.5 W
- 200 W
- 100 W
- 25 W
- 50 W
**Analysis:**
To determine the new rate of heat conduction when both the length and diameter of the cylindrical bar are doubled, we need to understand the relationship given by Fourier's Law of Heat Conduction:
\[ Q = \frac{KA \Delta T}{L} \]
where:
- \( Q \) is the rate of heat transfer,
- \( K \) is the thermal conductivity of the material,
- \( A \) is the cross-sectional area of the bar,
- \( \Delta T \) is the temperature difference between the hot and cold reservoirs,
- \( L \) is the length of the bar.
When the diameter (d) and the length (L) of the bar are both doubled:
1. **Cross-sectional Area (A):**
The area A is proportional to the square of the diameter.
\[ A_{\text{new}} = \pi \left(\frac{2d}{2}\right)^2 = 4 \cdot A \]
2. **Length (L):**
The new length \( L_{\text{new}} = 2L \)
Substituting these changes into Fourier's Law:
\[ Q_{\text{new}} = \frac{K (4A) \Delta T}{2L} = 2 \cdot \frac{KA \Delta T}{L} \]
Thus, the new rate of heat transfer \( Q_{\text{new}} \) is:
\[ Q_{\text{new}} = 2 \cdot Q = 2 \cdot 25 \text{ W} = 50 \text{ W} \]
**Answer:**
- 50 W](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec99e2ce-f146-4dfd-8358-2828b726596b%2Fdbbec449-58f8-4d9a-9ae2-d89cd03b9ce4%2Ft0kvn7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both the length and the diameter of this bar are doubled, the rate at which it will conduct heat between these reservoirs will be:
- 12.5 W
- 200 W
- 100 W
- 25 W
- 50 W
**Analysis:**
To determine the new rate of heat conduction when both the length and diameter of the cylindrical bar are doubled, we need to understand the relationship given by Fourier's Law of Heat Conduction:
\[ Q = \frac{KA \Delta T}{L} \]
where:
- \( Q \) is the rate of heat transfer,
- \( K \) is the thermal conductivity of the material,
- \( A \) is the cross-sectional area of the bar,
- \( \Delta T \) is the temperature difference between the hot and cold reservoirs,
- \( L \) is the length of the bar.
When the diameter (d) and the length (L) of the bar are both doubled:
1. **Cross-sectional Area (A):**
The area A is proportional to the square of the diameter.
\[ A_{\text{new}} = \pi \left(\frac{2d}{2}\right)^2 = 4 \cdot A \]
2. **Length (L):**
The new length \( L_{\text{new}} = 2L \)
Substituting these changes into Fourier's Law:
\[ Q_{\text{new}} = \frac{K (4A) \Delta T}{2L} = 2 \cdot \frac{KA \Delta T}{L} \]
Thus, the new rate of heat transfer \( Q_{\text{new}} \) is:
\[ Q_{\text{new}} = 2 \cdot Q = 2 \cdot 25 \text{ W} = 50 \text{ W} \]
**Answer:**
- 50 W
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