Question

Calculate the temperature at which the resistance of a conductor becomes 20% more than its resistance at $27^\circ\text{C}$. The value of the temperature coefficient of resistance of the material of conductor is $2.0 \times 10^{-4} ^\circ\text{C}^{-1}$.

Hint:

A "$20\%$ more" phrase immediately translates to a multiplier of $1.20$. Be extremely careful with scientific notation division to avoid off-by-ten calculation errors.

Answer 1

Concept:
• For standard solid metallic conductors, the intrinsic electrical resistance predictably and consistently increases as the ambient temperature increases.

• This specific thermal dependence over moderate temperature ranges is mathematically modeled by an established linear empirical approximation formula.

• The formula is rigorously given as $R_T = R_0 [1 + \alpha (T - T_0)]$, where $R_T$ is the new resistance at the target temperature $T$, $R_0$ is the known baseline resistance at reference temperature $T_0$, and $\alpha$ is the material's specific temperature coefficient of resistance.

Step 1: Extract and define all known parameters
The baseline reference temperature is explicitly provided as $T_0 = 27^\circ\text{C}$.
Let the initial resistance strictly at this reference temperature be denoted as $R_0$.
The material's specific temperature coefficient is provided as $\alpha = 2.0 \times 10^{-4} ^\circ\text{C}^{-1}$.
The problem firmly dictates that the final resistance $R_T$ at the unknown target temperature $T$ must become exactly 20% more than the baseline resistance $R_0$.
Mathematically, a 20% increase is cleanly formulated as:
\[ R_T = R_0 + 0.20 R_0 = 1.20 R_0 \]

Step 2: Set up the temperature dependence equation
We systematically substitute our defined parameters directly into the standard linear resistance formula:
\[ R_T = R_0 [1 + \alpha (T - T_0)] \]
\[ 1.20 R_0 = R_0 [1 + \alpha (T - 27)] \]

Step 3: Solve the algebra for the temperature change
Since the baseline resistance $R_0$ appears as a strict multiplier on both sides of the equation, we can cleanly and immediately cancel it out completely:
\[ 1.20 = 1 + \alpha (T - 27) \]
Subtract 1 from both sides to elegantly isolate the term containing the temperature difference:
\[ 1.20 - 1 = \alpha (T - 27) \]
\[ 0.20 = \alpha (T - 27) \]
Now, carefully substitute the provided numerical value for the temperature coefficient $\alpha$ into the equation:
\[ 0.20 = (2.0 \times 10^{-4}) \times (T - 27) \]
Vigorously rearrange the equation to solve for the explicit temperature difference $(T - 27)$:
\[ T - 27 = \frac{0.20}{2.0 \times 10^{-4}} \]

Step 4: Execute the final numerical calculation
To effectively handle the messy scientific notation in the denominator, multiply both numerator and denominator aggressively by $10^4$:
\[ T - 27 = \frac{0.20 \times 10^4}{2.0} \]
\[ T - 27 = \frac{2000}{2.0} \]
\[ T - 27 = 1000 \]
Finally, cleanly isolate the target temperature $T$ by shifting the reference temperature across:
\[ T = 1000 + 27 \]
\[ T = 1027^\circ\text{C} \]

Step 5: Conclusion
The conductor must be heated to a substantial temperature of exactly $1027^\circ\text{C}$ to experience a 20% increase in its electrical resistance.