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#temperature effect

22 public questions tagged with this topic.

At absolute zero (0 K), an intrinsic semiconductor behaves as:

**Semiconductor properties** distinguish from conductors and insulators by temperature dependence and doping response. At 0 K intrinsic acts as insulator, conductivity due to thermally generated electron-hole pairs, number of outer electrons 4 for Si/Ge forming covalent bonds, each atom shares electrons, crystal with N atoms has 4N valence electrons, 2N bonds. At 0 K, no thermal energy is available to excite electrons across the energy gap, so an intrinsic semiconductor has no free carriers and acts like an insulator. Substituting values gives Insulator, which matches expected behaviour for th

Ref: NCERT > Physics Book > Electronic Devices > Semiconductors, Types and Energy Bands

What happens to the drift velocity of electrons in a conductor if the conductor’s temperature increases while the applie

**Potentiometer** measures potential difference without drawing current, using null deflection, principle V ∝ l, l balance length, accurate because no I r drop. Potential drop across resistor V = I R arises because electric field does work on charges, energy converted to heat, maintaining E = -dV/dx along wire. Drift velocity v_d = e E tau / m . As temperature increases, tau (relaxation time) decreases due to more collisions, reducing v_d if E is constant. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R,

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

Why does a filament lamp deviate from Ohm’s law as current increases?

**Conductivity** σ = 1/ρ = n e² τ/m (S/m), τ relaxation time, m electron mass. Resistivity deviation at high fields occurs when τ depends on E or n changes due to impact ionization, breaking Ohm's law, seen in varistors, gas discharge. In a filament lamp, increased current raises the filament’s temperature, increasing its resistance due to a positive temperature coefficient. This makes the V -versus- I relationship non-linear, violating Ohm’s law. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R, evaluation yields Resista

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

Why does a semiconductor’s resistance decrease with increasing temperature, unlike a metal?

**Current and drift relation** I = n e A v_d shows current proportional to drift velocity and area. For A=6×10⁻⁷ m², I=1.8 A, n=8.5×10²⁸ m⁻³, v_d =1.8/(8.5×10²⁸×1.6×10⁻¹⁹×6×10⁻⁷)=2.2×10⁻⁴ m/s, illustrating small drift speed even for ampere currents. In semiconductors, as temperature rises, more electrons are thermally excited from the valence band to the conduction band, increasing the number of charge carriers ( n ). Since rho = m / (n e² tau) , a significant increase in n outweighs the decrease in tau , reducing rho and thus resistance. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kir

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

A conductor has a resistivity of \( 6 \times 10^{-8} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\te

**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Use: rho_t = rho₀ [1 + α (T - T₀)] . Substitute: rho_t = 6 × 10⁻⁸ [1 + 4 × 10⁻³ (90 - 20)] . Calculate: rho_t = 6 × 10⁻⁸ [1 + 0.28] = 6 × 10⁻⁸ × 1.28 = 7.68 × 10⁻⁸ Ω m . Applying I = n e A v_d, R = ρ l/A, R_t

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

A wire has a resistance of \( 25 \, \Omega \) at \( 20^\circ \text{C} \) and \( 27.5 \, \Omega \) at \( 90^\circ \text{C

**Temperature dependence** of resistance R_t = R₀[1+α(T-T₀)], α temperature coefficient (per °C), R₀ resistance at T₀ (Ω). For metals α positive ≈10⁻³ /°C, resistance increases with temperature because τ decreases due to increased phonon scattering, n nearly constant. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: 27.5 = 25 [1 + α (90 - 20)] . Solve: 27.5 = 25 + 1750α ⇒ 1750α = 2.5 ⇒ α = (2.5/1750) ≈ 1.43 × 10⁻³ °C⁻¹ . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

A wire has a resistance of \( 15 \, \Omega \) at \( 20^\circ \text{C} \) and \( 16.5 \, \Omega \) at \( 70^\circ \text{C

**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: 16.5 = 15 [1 + α (70 - 20)] . Solve: 16.5 = 15 + 750α ⇒ 750α = 1.5 ⇒ α = (1.5/750) = 2 × 10⁻³ °C⁻¹ . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT],

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

In a metallic conductor, if the temperature increases significantly, what happens to the conductivity of the material?

**Resistivity temperature variation** ρ_t = ρ₀[1+α(T-T₀)], α ≈4×10⁻³ /°C for copper, 1.7×10⁻⁴ /°C for nichrome. Given R=60 Ω at 30°C, α=1.7×10⁻⁴ /°C, T=330°C, ΔT=300°C, R_t=60[1+1.7×10⁻⁴×300]=60×1.051=63.06 Ω, modest increase for nichrome due to small α. Conductivity ( sigma = 1 / rho ) is inversely proportional to resistivity ( rho = m / (n e² tau) ). As temperature increases, the relaxation time ( tau ) decreases due to more frequent collisions, increasing rho and thus decreasing sigma . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε -

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

Which wave property is most affected when a sound wave encounters a change in the medium’s temperature?

**Doppler effect** describes apparent frequency shift due to relative motion between source and observer, f' = f·v/(v ∓ v_s) for source motion, f' = f·(v ± v_o)/v for observer motion, upper signs for approach increasing observed frequency. Motion towards observer compresses wavelength raising f'. Speed of sound in a gas increases with temperature ( v ∝ √(T) ), significantly altering its propagation, while frequency remains source-dependent and amplitude may vary less directly. Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Speed, illustrating

Ref: NCERT > Physics Book > Waves > Doppler Effect

Why does the speed of sound increase with temperature in a gas?

**Longitudinal vibrations in rods** clamped at middle have fundamental with node at clamp and antinodes at ends, f₁ = v/(2L), v speed of sound in material (m/s). This relation allows v extraction from measured f₁ and length L, e.g., v = 2L·f₁. Speed of sound in a gas is v = √((gamma P/rho)) , and since P/rho ∝ T (ideal gas law), higher temperature increases molecular velocity, thus increasing v . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Molecular velocity increases, illustrating frequency-length-speed interdependence and quantization by

Ref: NCERT > Physics Book > Waves > Sound Waves, Reflection and Characteristics

Why is the specific heat capacity of a substance temperature-dependent?

**Isobaric and isothermal** are fundamental thermodynamic processes, isobaric P constant horizontal line on P-V diagram, isothermal hyperbolic P = n R T/V, work equals area under curve, isothermal work larger than adiabatic for same volume change because pressure higher. Specific heat capacity varies with temperature because the energy required to raise the temperature of a substance depends on molecular interactions and vibrational modes, which change with temperature (e.g., water’s variation in Fig. 11.5). Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

The rms speed of nitrogen molecules is 516 m/s at 300 K. What will it be at 600 K? (Molecular mass of N₂ = 28 u)

**Ideal gas equation** combines Boyle, Charles, Avogadro laws, P V = N k_B T, N number of molecules, k_B Boltzmann constant, for 1 mole N_A=6.022×10²³, R = N_A k_B, enabling calculation of volume from P,T,n. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(v₂)/(516) = √((600)/(300)) = √(2) ≈ 1.414.v₂ = 516 × 1.414 ≈ 729 m/s . Substituting values gives 729 m/s, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation