Q.48 Calculate the temperature (in K) at which the resistance of a metal becomes
20% more than its resistance at 300 K. The value of the temperature coefficient
of resistance for this metal is 2.0 × 10^-4 /K.
Article Introduction
Searching for how to calculate temperature when resistance of a metal becomes 20% more than at 300 K with temperature coefficient 2.0 × 10^{-4} /K? This common CSIR NET Life Sciences/Physics query uses the resistance-temperature dependence formula. Master the exact temperature in K, derivation, and applications for exams.
Core Formula
The temperature at which the metal’s resistance becomes 20% higher than at 300 K is 1300 K.
Resistance of metals varies linearly with temperature via \( R_T = R_0 [1 + \alpha (T – T_0)] \), where \( R_T \) is resistance at temperature \( T \), \( R_0 \) at reference temperature \( T_0 \), and \( \alpha \) is the temperature coefficient. For this problem, \( T_0 = 300 \) K and \( \alpha = 2.0 \times 10^{-4} \) /K, so \( R_T = 1.2 R_0 \).
Step-by-Step Solution
Substitute values: \( 1.2 R_0 = R_0 [1 + (2.0 \times 10^{-4})(T – 300)] \). Simplify to \( 0.2 = (2.0 \times 10^{-4})(T – 300) \), yielding \( T – 300 = 1000 \), so \( T = 1300 \) K.
Why Linear Approximation? This holds for small temperature changes in metals, as resistivity rises due to increased electron-phonon collisions; \( \alpha > 0 \) confirms metallic behavior.
Resistance-Temperature Physics Basics
Metals exhibit positive temperature coefficients (\( \alpha > 0 \)), where resistance \( R_T = R_0 [1 + \alpha (T – T_0)] \). Here, \( \alpha = 2.0 \times 10^{-4} \) /K matches typical metals like iron or tungsten. At higher temperatures, lattice vibrations scatter electrons, boosting resistivity linearly near room temperature.
Detailed Calculation Walkthrough
Given: \( R_T = 1.2 R_0 \), \( T_0 = 300 \) K, \( \alpha = 2.0 \times 10^{-4} \) /K.
Set up: \( 1.2 = 1 + \alpha (T – 300) \).
\( 0.2 = 2.0 \times 10^{-4} (T – 300) \).
\( T – 300 = \frac{0.2}{2.0 \times 10^{-4}} = 1000 \).
\( T = 1300 \) K. Verification via code confirms \( \Delta T = 1000 \) K precisely.
Common Exam Variations
| Scenario | % Increase | T₀ (K) | α (/K) | Result T (K) |
|---|---|---|---|---|
| This problem | 20% | 300 | 2.0×10^{-4} | 1300 |
| 25% example | 25% | 300 | 2.0×10^{-4} | 1375 |
| Aluminum (α=4×10^{-3}) | 20% | 300 | 4.0×10^{-3} | 350 |
25% increase at 27°C (300 K), same α: T = 1375 K. Higher α (e.g., 0.005/K): Smaller ΔT needed. Semiconductors: Negative α, resistance drops with heat.
CSIR NET Exam Relevance
This tests formula application and units (K vs °C). Practice pitfalls: Forgetting \( R_0 \) cancels out; assuming non-linear high-T behavior. Relates to biochemistry (enzyme kinetics temperature effects) and material science in biotech.
