Zaymiey

⚛️ Physics  ·  Thermal Properties of Matter  ·  NEET & JEE

A composite slab is made of two materials of equal thickness with thermal conductivities K1 and K2 placed in series (heat flows perpendicular to the layers). The effective thermal conductivity of the slab is:

Answer: 2K1K2/(K1 + K2).

  • A (K1 + K2)/2
  • B 2K1K2/(K1 + K2)
  • C K1K2/(K1+K2)
  • D K1 + K2

Correct answer: B. 2K1K2/(K1 + K2)

Explanation: For slabs of equal thickness in series, thermal resistances add (R = L/KA), giving an effective conductivity analogous to resistors in series: K<sub>eff</sub> = 2K1K2/(K1+K2).

Heating Curve: Ice → Water → SteamHeat added (Q)T (°C)ice warms (-ve→0°C)melting (0°C, latent heat)water warms (0→100°C)boiling (100°C, latent heat)FLAT regions = phase change (all heat goes into breaking bonds, NOT raising temperature)

During a phase change (melting or boiling), temperature stays constant while heat is absorbed entirely as latent heat (Q=mL); temperature only rises again once the substance is fully in its new phase.

Concept context

Temperature scales, thermal expansion, calorimetry, and the three modes of heat transfer including radiation laws.

Read the full Thermal Properties of Matter notes →