NEB Class 11 • Physics • Heat Transfer

Rate of Heat Flow: NEB Class 11 Physics Guide

Separate total transferred energy from transfer rate. Use temperature difference, area, path length and material properties to explain insulation, conductors and steady heat flow.

P = Q/t and Fourier conduction model
Thermal conductivity, resistance and scaling
Convection, radiation, applications and examples

Heat-transfer rate depends on the pathway as well as the temperature difference.

Curriculum and model

What Rate of Heat Flow means

The CDC Physics Grade 11 page and secondary curriculum define the Nepal learning boundary. This guide uses the steady-conduction equation described in OpenStax Conduction and compares it with the broader heat-transfer methods.

Rate answers “how quickly?” Quantity of heat answers “how much?” A wall may transfer a large total energy during a long winter while having a modest instantaneous rate. Always include time when converting between P and Q.

Energy per time

Power, heat-flow direction and equilibrium

P = Q/t has SI unit J s⁻¹ = W. Net thermal energy flows spontaneously from higher to lower temperature. When the two sides have equal temperature, the net rate becomes zero even though microscopic energy exchange continues in both directions.

Energy from a known rate

A roof loses thermal energy at 600 W for 30 minutes. Q = Pt = 600×1800 = 1.08×10⁶ J. Convert minutes to seconds because watts are joules per second.

Average versus instantaneous rate

If 90 kJ transfers in 5 minutes, average P = 90,000/300 = 300 W. The instantaneous rate may have changed as temperatures approached equilibrium; the result is only the interval average.

Fourier model

Steady conduction through a slab

For a uniform slab of thickness L, area A and thermal conductivity k with face temperatures T_h and T_c, the magnitude of steady heat-flow rate is P = kA(T_h−T_c)/L. Conductivity k has unit W m⁻¹K⁻¹.

The equation assumes approximately steady one-dimensional flow, constant k, uniform cross-section, known surface temperatures and negligible side losses. It does not directly describe the early transient after one face is suddenly heated.

Factor	Change	Rate response
k	doubles	P doubles
A	triples	P triples
ΔT	halves	P halves
L	doubles	P halves

Worked wall example

A wall has k=0.80 W m⁻¹K⁻¹, area 10 m², thickness 0.20 m and temperature difference 15 K. P=0.80×10×15/0.20=600 W.

A useful comparison

Thermal resistance and composite layers

Define slab thermal resistance R_th = L/(kA), unit K W⁻¹. Then P = ΔT/R_th. For layers in series carrying the same steady rate, resistances add: R_total = ΣLᵢ/(kᵢA), and P = ΔT_total/R_total.

Two-layer wall

For equal area 5 m², layer 1 has L=0.10 m, k=0.50; layer 2 has L=0.05 m, k=0.10. R₁=0.10/(0.50×5)=0.04 K W⁻¹; R₂=0.05/(0.10×5)=0.10 K W⁻¹. Total R=0.14 K W⁻¹. For ΔT=14 K, P=100 W. The thin low-k layer contributes most resistance.

This is analogous to electrical resistance but the mechanisms are not identical. State the thermal definitions instead of copying circuit formulas without justification.

Applications and scaling

Worked Rate of Heat Flow examples

Window comparison

Two windows have the same k, thickness and ΔT; one has twice the area. Its conductive rate is twice as large. Total energy loss also depends on duration.

Insulation layer

If a uniform insulation thickness rises from 3 cm to 6 cm with other factors fixed, the ideal slab rate halves. In practice, air leakage and thermal bridges can prevent the whole-building loss from halving.

Metal spoon and wooden handle

Metal often has much larger k than wood, so it transfers energy from hot tea more quickly. The sensation depends on rate, not necessarily a higher initial metal temperature.

Find conductivity

A slab transfers 240 W through area 2.0 m² and thickness 0.050 m under ΔT=12 K. k=PL/(AΔT)=240×0.050/(2×12)=0.50 W m⁻¹K⁻¹.

Beyond solid conduction

Convection and radiation also control rate

Convection transfers energy through bulk fluid motion and can be natural or forced. A fan can increase heat-transfer rate near a warm surface by replacing warmed boundary air. Radiation transfers energy by electromagnetic waves and does not require matter; surface temperature, area and emissivity matter strongly.

Real Kathmandu rooms combine conduction through walls and windows, convection by air and radiation between surfaces and outdoors. The NASA thermal energy transfer resource supports a multi-mechanism classroom model. Do not attribute every loss to conduction merely because a wall is present.

Conduction

Dominant inside a solid path; controlled by k and geometry.

Convection

Bulk fluid movement; wind and fans can change rate.

Radiation

Electromagnetic transfer; no material medium required.

Practical investigation

Compare thermal conductivities

Use equal-length, equal-area rods of different materials with one end heated under teacher supervision. Measure temperature at comparable positions and times, or compare a calibrated steady transfer setup. Keep geometry, contact and energy input controlled so material is the main changed variable.

A quick temperature rise at a sensor suggests greater effective transfer, but heat capacity also affects transient warming. A stronger conductivity comparison uses steady state or a model that includes mass and specific heat. Never touch heated rods; use guards, tongs and approved temperatures.

Variable	Control
Length and area	Use matched dimensions
Contact with heater	Use same pressure/interface method
Sensor position	Measure from the same reference point
Side losses	Use comparable insulation and airflow
Initial temperature	Allow all rods to equilibrate first

Problem routine

Seven steps for rate questions

Identify hot and cold boundaries.
Decide whether the requested quantity is Q or P.
State the heat-transfer mechanism.
Draw area perpendicular to flow and thickness along flow.
Check steady, uniform and one-dimensional assumptions.
Substitute SI units and calculate.
Test scaling, direction and watts-versus-joules.

Common failures are using wall face length instead of thickness, confusing conductivity k with heat capacity c, leaving minutes unconverted, adding layer conductivities directly, and reporting joules when the question asks watts. Review Quantity of Heat to connect P with total Q.

For online or physical NEB tuition, call 9846662070 with your diagram and scaling check.

Practice tasks

Rate of Heat Flow checkpoint

Distinguish 500 J from 500 W.
Derive the unit of k.
Predict rate when area triples and thickness doubles.
Solve one uniform-wall problem.
Find k from measured data.
Combine two slab resistances in series.
Classify conduction, convection and radiation in a solar water heater.
Evaluate a rod-comparison experiment.

Exam tip: Draw the direction arrow first. Mark A across the arrow and L along it; this prevents the most common geometry reversal.

Frequently asked questions

Questions about Rate of Heat Flow

What is the SI unit of heat-flow rate?

The watt, W, equal to one joule per second.

What is the SI unit of thermal conductivity?

W m−1 K−1, obtained from P=kAΔT/L.

Why does a thicker wall reduce steady conduction?

The energy must cross a longer path, so thermal resistance L/(kA) increases.

Can I add conductivities of wall layers?

No. For series layers with common area and steady rate, add thermal resistances L/(kA), then divide total temperature difference by total resistance.

Why can metal and wood feel different at the same temperature?

Metal usually transfers energy to or from skin faster because its thermal conductivity is higher.

Where can I get NEB Rate of Heat Flow help?

For current online or physical tuition options, call 9846662070 and confirm timetable, class mode, teacher availability and fees.

References and next steps

Sources and related study guides

Use the Heat & Temperature Study Guide to interleave conduction with calorimetry, expansion and phase change. Curriculum scope and sources were checked on 2 August 2026; follow current CDC, NEB and college instructions if requirements change.

Advanced comparison

Temperature profile and interface reasoning

In steady one-dimensional conduction through two layers in series, the same rate crosses each layer; otherwise energy would accumulate at their interface. The temperature drop across a layer is ΔTᵢ = PRᵢ, so the larger thermal resistance has the larger temperature drop. This gives a physical way to check composite-wall algebra.

Interface temperature

Two layers carry P=100 W with resistances 0.04 and 0.10 K W⁻¹. Their drops are 4 K and 10 K. If the hot face is 30°C and layer 1 comes first, the interface is 26°C and the cold face is 16°C. Drops add to the total 14 K.

Use the Physical Quantities guide to verify that K W⁻¹ times W gives K. Then sketch temperature against position: the slope magnitude is steeper in the lower-conductivity layer when area and rate are common. The graph is piecewise linear only under the ideal constant-k steady model.

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