The ends of a cylinder steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal`//` s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?

Options

A. 5 cal `//`s
B. 10 cal `//` s
C. 20 cal `//` s
D. 40 cal `//` s

Answer

To solve the problem, we will use the formula for heat conduction, which is given by Fourier's law of heat conduction. The rate of heat transfer (Q) through a material is given by:

\[ Q = \frac{k \cdot A \cdot \Delta T}{L} \]

Where:
- \( Q \) = rate of heat transfer (in calories per second)
- \( k \) = thermal conductivity of the material (constant for steel)
- \( A \) = cross-sectional area of the rod
- \( \Delta T \) = temperature difference between the two ends
- \( L \) = length of the rod

### Step 1: Identify the parameters of the original rod
- The original rod conducts heat at a rate of \( Q_1 = 10 \) cal/s.
- Let the length of the original rod be \( L \).
- The diameter of the original rod is \( d \).
- The cross-sectional area \( A_1 \) of the original rod is given by:
\[ A_1 = \frac{\pi d^2}{4} \]

### Step 2: Determine the parameters of the new rod
- The new rod is twice as long, so its length \( L_2 = 2L \).
- The diameter of the new rod is twice that of the original, so \( d_2 = 2d \).
- The cross-sectional area \( A_2 \) of the new rod is given by:
\[ A_2 = \frac{\pi (2d)^2}{4} = \frac{\pi (4d^2)}{4} = \pi d^2 = 4A_1 \]

### Step 3: Calculate the rate of heat transfer for the new rod
Using the heat conduction formula for the new rod:
\[ Q_2 = \frac{k \cdot A_2 \cdot \Delta T}{L_2} \]

Substituting the values:
- \( A_2 = 4A_1 \)
- \( L_2 = 2L \)

So we have:
\[ Q_2 = \frac{k \cdot (4A_1) \cdot \Delta T}{2L} \]

### Step 4: Simplify the equation
We can simplify the equation:
\[ Q_2 = \frac{4k \cdot A_1 \cdot \Delta T}{2L} = 2 \cdot \frac{k \cdot A_1 \cdot \Delta T}{L} = 2Q_1 \]

Since \( Q_1 = 10 \) cal/s, we find:
\[ Q_2 = 2 \cdot 10 = 20 \text{ cal/s} \]

### Conclusion
The rate at which the new steel rod conducts heat is \( Q_2 = 20 \) cal/s.

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