Residential building codes typically require the use of 12-gauge copper wire (diameter 0.2053 cm) for wiring receptacles. Such circuits carry currents as large as 20 A. If a wire of smaller diameter (with a higher gauge number) carried that much current, the wire could rise to a high temperature and cause a fire.

(a) Calculate the rate at which internal energy is produced in 2.10 m of 12-gauge copper wire carrying a current of 20.0 A.
 W
(b) Repeat the calculation for an aluminum wire.
 WWould a 12-gauge aluminum wire be as safe as a copper wire?

### Part (a) - **Copper Wire**

To calculate the rate at which internal energy is produced in a 12-gauge copper wire, we use the formula for **power dissipation due to resistance**:

\[
P = I^2 R
\]

Where:
- \( P \) is the power dissipation (rate of internal energy production) in watts (W),
- \( I \) is the current in amperes (A),
- \( R \) is the resistance of the wire in ohms (Ω).

The resistance \( R \) of the wire can be calculated using the formula:

\[
R = \rho \frac{L}{A}
\]

Where:
- \( \rho \) is the resistivity of the material (copper),
- \( L \) is the length of the wire (2.10 m),
- \( A \) is the cross-sectional area of the wire.

#### Step 1: Cross-sectional Area \( A \)
The wire's diameter is given as 0.2053 cm, which we convert to meters:

\[
d = 0.2053 \text{ cm} = 0.002053 \text{ m}
\]

The cross-sectional area \( A \) of a circular wire is:

\[
A = \pi \left( \frac{d}{2} \right)^2
\]

\[
A = \pi \left( \frac{0.002053}{2} \right)^2
\]

\[
A \approx 3.31 \times 10^{-6} \text{ m}^2
\]

#### Step 2: Resistance \( R \)
The resistivity \( \rho \) of copper is \( 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \). Now we can calculate the resistance:

\[
R = \rho \frac{L}{A}
\]

\[
R = (1.68 \times 10^{-8}) \frac{2.10}{3.31 \times 10^{-6}}
\]

\[
R \approx 0.0107 \, \Omega
\]

#### Step 3: Power Dissipation \( P \)
Now using the current \( I = 20.0 \, A \), we calculate the power:

\[
P = I^2 R
\]

\[
P = (20.0)^2 \times 0.0107
\]

\[
P \approx 4.28 \, \text{W}
\]

So, the rate at which internal energy is produced in the copper wire is approximately **4.28 W**.

---

### Part (b) - **Aluminum Wire**

For aluminum, the calculation is similar, but the resistivity of aluminum is different. The resistivity \( \rho \) of aluminum is \( 2.82 \times 10^{-8} \, \Omega \cdot \text{m} \).

#### Step 1: Resistance \( R \)
Using the same cross-sectional area \( A = 3.31 \times 10^{-6} \, \text{m}^2 \) and length \( L = 2.10 \, \text{m} \):

\[
R = \rho \frac{L}{A}
\]

\[
R = (2.82 \times 10^{-8}) \frac{2.10}{3.31 \times 10^{-6}}
\]

\[
R \approx 0.0179 \, \Omega
\]

#### Step 2: Power Dissipation \( P \)
Now using the current \( I = 20.0 \, A \):

\[
P = I^2 R
\]

\[
P = (20.0)^2 \times 0.0179
\]

\[
P \approx 7.16 \, \text{W}
\]

So, the rate at which internal energy is produced in the aluminum wire is approximately **7.16 W**.

---

### Would a 12-gauge aluminum wire be as safe as a copper wire?

No, a 12-gauge aluminum wire would not be as safe as a copper wire. As we can see from the calculations, the power dissipation (and therefore the heat produced) in the aluminum wire is much higher than in the copper wire (7.16 W vs. 4.28 W). This increased heat could lead to a higher risk of fire or damage in the aluminum wire under the same current load.

---

Would you like further details on any of these calculations?

### Related Questions:
1. What are the advantages of using copper over aluminum for wiring?
2. How does the resistivity of a material affect its suitability for electrical wiring?
3. What safety measures can be implemented when using aluminum wires in residential buildings?
4. What happens to the resistance of wires as their temperature increases?
5. How does wire length affect the rate of energy loss in electrical circuits?

### Tip:
Always ensure the wire gauge and material are appropriate for the current to minimize the risk of overheating and potential fire hazards.

Residential building codes typically require the use of 12-gauge copper wire (diameter 0.2053 cm) for wiring receptacles. Such circuits carry currents as large as 20 A. If a wire of smaller diameter (with a higher gauge number) carried that much current, the wire could rise to a high temperature and cause a fire. (a) Calculate the rate at which internal energy is produced in 2.10 m of 12-gauge copper wire carrying a current of 20.0 A. (b) Repeat the calculation for an aluminum wire. Would a 12-gauge aluminum wire be as safe as a copper wire?

Explore how to calculate the rate of internal energy production in a 12-gauge copper wire and an aluminum wire carrying a current of 20 A. Learn the key physics concepts, including power dissipation, resistance, and Ohm's law, with detailed steps. Suitable for high school physics students.

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