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NEET-UG and JEE Main Physics · Class 12

Specific heat capacity by the method of mixtures — NEET & JEE Questions

Specific heat capacity by the method of mixtures is a topic in Experimental Skills, a Class 12 Physics chapter of the NEET-UG and JEE Main syllabus. TestBudy holds 14 published NEET-UG and JEE Main questions on Specific heat capacity by the method of mixtures, each with the marking scheme of the real paper.

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Specific heat capacity by the method of mixtures MCQs with solutions

10 of the 14 published questions on this topic, exactly as they appear in a test. Answers and worked solutions come with your result rather than on this page.

  1. Question 1 · difficulty L1 · recall

    Specific heat capacity is the heat required to raise

    • A. unit mass of a substance by one kelvin
    • B. any mass by 100 K
    • C. unit volume by one second
    • D. temperature without energy input

  2. Question 2 · difficulty L1 · understanding

    The method of mixtures is based mainly on

    • A. conservation of electric charge only
    • B. conservation of energy
    • C. Snell's law
    • D. Faraday induction

  3. Question 3 · difficulty L2 · understanding

    For an isolated hot object placed in cooler water, thermal equilibrium is reached when

    • A. both reach 0°C
    • B. the hotter object has zero specific heat
    • C. they have the same final temperature
    • D. their masses become equal

  4. Question 4 · difficulty L2 · application

    If heat loss to surroundings is negligible, heat lost by hot body equals

    • A. zero regardless of temperature
    • B. only heat gained by air
    • C. mass lost by the hot body
    • D. heat gained by water plus calorimeter

  5. Question 5 · difficulty L2 · understanding

    The method of mixtures is based on the principle that, if heat loss to surroundings is negligible,

    • A. heat lost by the hotter body equals heat gained by the colder body and calorimeter
    • B. the hotter body loses no heat
    • C. both bodies must have equal masses
    • D. temperature change is independent of specific heat

  6. Question 6 · difficulty L3 · application

    A 0.10 kg metal of specific heat 400 J kg^-1 K^-1 cools by 50 K. Heat released is

    • A. 2000 J
    • B. 200 J
    • C. 20,000 J
    • D. 8 J

  7. Question 7 · difficulty L3 · application

    Why is the calorimeter often given a water equivalent or heat capacity correction?

    • A. It has no mass
    • B. The calorimeter itself absorbs heat
    • C. Water cannot absorb heat
    • D. It changes gravitational acceleration

  8. Question 8 · difficulty L3 · application

    A 100 g metal at 80°C is placed in 100 g water at 20°C. The final temperature is 30°C. Neglect the calorimeter and take water's specific heat as 4.2 J g−1^{-1} K−1^{-1}. What is the metal's specific heat?

    • A. 0.42 J g−1^{-1} K−1^{-1}
    • B. 0.84 J g−1^{-1} K−1^{-1}
    • C. 4.2 J g−1^{-1} K−1^{-1}
    • D. 8.4 J g−1^{-1} K−1^{-1}

  9. Question 9 · difficulty L3 · application

    A 50 g metal sample of specific heat 0.40 J g−1^{-1} K−1^{-1} cools by 30 K. How much heat does it release?

    • A. 60 J
    • B. 240 J
    • C. 600 J
    • D. 1500 J

  10. Question 10 · difficulty L4 · analysis

    If some heat escapes to the surroundings during mixing but is ignored, the final temperature will be

    • A. higher than the hot sample's initial temperature
    • B. unchanged exactly
    • C. lower than the ideal isolated value when a hot sample heats cooler water
    • D. always 0°C

Common mistakes in Specific heat capacity by the method of mixtures

Recorded against questions in this topic that are not shown above, so none of these points at an answer you can see on this page.

  • Ignores heat absorbed by the calorimeter and therefore overestimates the measured heat available to the water.

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