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⚛️ Physics  ·  Thermal Properties of Matter  ·  NEET & JEE

A calorimeter of negligible heat capacity contains 200 g of water at 25°C. 50 g of ice at 0°C is added. Given latent heat of fusion = 336 J/g and specific heat of water = 4.2 J/(g·K), the final temperature of the mixture is approximately:

Answer: 5°C.

  • A 0°C (some ice remains)
  • B 5°C
  • C 10°C
  • D 15°C

Correct answer: B. 5°C

Explanation: Heat released by water cooling to 0°C = 200×4.2×25 = 21000 J. Heat needed to melt all the ice = 50×336 = 16800 J. Remaining heat = 4200 J raises the combined 250 g of water: 4200 = 250×4.2×ΔT, giving ΔT = 4°C, so final temperature is approximately 4 to 5°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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