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⚛️ Physics  ·  Thermal Properties of Matter  ·  NEET & JEE

Equal masses of two liquids A (specific heat 2000 J/kgK) and B (specific heat 4000 J/kgK) are mixed; A is at 80°C and B is at 20°C. What is the equilibrium temperature (assuming no heat loss)?

Answer: 33.3°C.

  • A 40°C
  • B 33.3°C
  • C 46.6°C
  • D 50°C

Correct answer: B. 33.3°C

Explanation: Heat lost by A = heat gained by B: m×2000×(80-T) = m×4000×(T-20). Solving: 2000(80-T)=4000(T-20) → 160000-2000T=4000T-80000 → 240000=6000T → T=40°C. Correct equilibrium temperature is 40°C.

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.

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