To solve the problem, we need to find the ratio of heat produced in two aluminum wires of different diameters when they are connected in series and the same current is passed through them. Here’s a step-by-step breakdown of the solution:
### Step 1: Understand the relationship between resistance and heat produced
When two resistors (or wires in this case) are connected in series, the heat produced in each wire can be calculated using the formula:
\[ P = I^2 R \]
where \( P \) is the power (heat produced per unit time), \( I \) is the current, and \( R \) is the resistance of the wire.
### Step 2: Calculate the resistance of each wire
The resistance \( R \) of a wire is given by:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area.
For the two wires:
- Wire 1 (diameter \( D_1 = 1 \, \text{mm} \)):
\[ A_1 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} (1 \, \text{mm})^2 \]
- Wire 2 (diameter \( D_2 = 2 \, \text{mm} \)):
\[ A_2 = \frac{\pi}{4} D_2^2 = \frac{\pi}{4} (2 \, \text{mm})^2 \]
Thus, the resistances are:
- \( R_1 = \frac{\rho L_1}{A_1} = \frac{\rho L_1}{\frac{\pi}{4} (1 \, \text{mm})^2} = \frac{4 \rho L_1}{\pi (1 \, \text{mm})^2} \)
- \( R_2 = \frac{\rho L_2}{A_2} = \frac{\rho L_2}{\frac{\pi}{4} (2 \, \text{mm})^2} = \frac{4 \rho L_2}{\pi (2 \, \text{mm})^2} \)
### Step 3: Set up the ratio of resistances
The ratio of the resistances \( R_1 \) and \( R_2 \) is:
\[
\frac{R_1}{R_2} = \frac{4 \rho L_1 / (\pi (1 \, \text{mm})^2)}{4 \rho L_2 / (\pi (2 \, \text{mm})^2)}
\]
This simplifies to:
\[
\frac{R_1}{R_2} = \frac{L_1}{L_2} \cdot \frac{(2 \, \text{mm})^2}{(1 \, \text{mm})^2} = \frac{L_1}{L_2} \cdot 4
\]
### Step 4: Relate lengths of wires to their volumes
Since both wires are made from the same mass of aluminum, their volumes must be equal:
\[
V_1 = V_2 \implies A_1 L_1 = A_2 L_2
\]
Substituting the areas:
\[
\frac{\pi}{4} (1 \, \text{mm})^2 L_1 = \frac{\pi}{4} (2 \, \text{mm})^2 L_2
\]
This simplifies to:
\[
(1 \, \text{mm})^2 L_1 = (2 \, \text{mm})^2 L_2 \implies L_1 = 4 L_2
\]
### Step 5: Substitute back to find the ratio of heat produced
Now substituting \( L_1 = 4 L_2 \) into the resistance ratio:
\[
\frac{R_1}{R_2} = \frac{4 L_2}{L_2} \cdot 4 = 16
\]
### Step 6: Calculate the ratio of heat produced
Using the power formula:
\[
\frac{P_1}{P_2} = \frac{I^2 R_1}{I^2 R_2} = \frac{R_1}{R_2} = 16
\]
### Final Result
The ratio of heat produced in the two wires is:
\[
\text{Heat produced in Wire 1 : Heat produced in Wire 2} = 16 : 1
\]
