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Calorimetry & Thermal Properties — NEET Physics MCQs with Solutions
Free NEET Physics Calorimetry & Thermal Properties MCQs with step-by-step solutions covering Specific Heat & Heat Capacity, Principle of Calorimetry & Mixtures, Latent Heat & Phase Transitions, Ice-Water-Steam Mixing Problems. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Specific Heat & Heat Capacity · easy · theory
The specific heat capacity of a substance is the heat required to raise the temperature of:
A. One mole of the substance by 1°C
B. Unit mass of the substance by 1°C (or 1 K) ✓ Correct
C. The whole body by 1°C
D. Unit volume by 1°C
Solution: c = Q/(mΔT): heat per unit mass per unit temperature rise. Units: cal/g°C or J/kg·K.
Q2 — Specific Heat & Heat Capacity · easy · theory
The heat capacity (thermal capacity) C of a body is related to its specific heat c by:
A. C = mc²
B. C = m/c
C. C = c/m
D. C = mc ✓ Correct
Solution: Heat capacity = heat to raise the WHOLE body by 1°C: C = mc (units cal/°C or J/K).
Q3 — Specific Heat & Heat Capacity · easy · theory
The molar heat capacity C_m of a substance of molar mass M and specific heat c is:
A. C_m = M/c
B. C_m = c
C. C_m = c/M
D. C_m = M·c ✓ Correct
Solution: Molar heat capacity = heat per mole per degree = (mass of one mole) × (specific heat) = Mc.
Q4 — Specific Heat & Heat Capacity · easy · theory
The SI unit of specific heat capacity is:
A. J·kg/K
B. cal/g°C
C. J/kg·K ✓ Correct
D. J/K
Solution: SI: joule per kilogram per kelvin (J/kg·K). cal/g°C is the practical (CGS-style) unit; 1 cal/g°C = 4200 J/kg·K.
Q5 — Specific Heat & Heat Capacity · easy · theory
Among common substances, water has:
A. Zero specific heat
B. One of the highest specific heat capacities (1 cal/g°C) ✓ Correct
C. The lowest specific heat
D. A specific heat equal to that of ice
Solution: Water's exceptionally high specific heat (1 cal/g°C = 4200 J/kg·K) is why it is used as a coolant and moderates climate. (Ice is about half: 0.5 cal/g°C.)
Q6 — Specific Heat & Heat Capacity · easy · numerical
The heat required to raise the temperature of 500 g of water from 20°C to 60°C is:
A. 20000 cal ✓ Correct
B. 10000 cal
C. 30000 cal
D. 40000 cal
Solution: Q = mcΔT = 500 × 1 × (60 − 20) = 20000 cal = 20 kcal.
Q7 — Specific Heat & Heat Capacity · easy · numerical
The heat needed to raise the temperature of 2 kg of water by 5 K (c_w = 4200 J/kg·K) is:
A. 4200 J
B. 8400 J
C. 21000 J
D. 42000 J ✓ Correct
Solution: Q = mcΔT = 2 × 4200 × 5 = 42000 J = 42 kJ.
Q8 — Specific Heat & Heat Capacity · easy · numerical
100 g of copper (c = 0.1 cal/g°C) is heated from 30°C to 80°C. The heat absorbed is:
A. 50 cal
B. 5000 cal
C. 500 cal ✓ Correct
D. 800 cal
Solution: Q = mcΔT = 100 × 0.1 × 50 = 500 cal.
Q9 — Specific Heat & Heat Capacity · easy · numerical
The heat capacity of 5 kg of a substance whose specific heat is 400 J/kg·K is:
A. 5000 J/K
B. 400 J/K
C. 80 J/K
D. 2000 J/K ✓ Correct
Solution: C = mc = 5 × 400 = 2000 J/K.
Q10 — Specific Heat & Heat Capacity · easy · numerical
A body of heat capacity 800 J/K absorbs 4000 J of heat. Its temperature rise is:
A. 8 K
B. 0.2 K
C. 5 K ✓ Correct
D. 2 K
Solution: ΔT = Q/C = 4000/800 = 5 K.
Q11 — Principle of Calorimetry & Mixtures · easy · theory
The principle of calorimetry states that, for an isolated system:
A. Heat lost by the hot body = heat gained by the cold body ✓ Correct
B. Heat always flows from cold to hot
C. The final temperature is the average of the initial temperatures
D. No heat is exchanged
Solution: Core equation: Heat lost = Heat gained (conservation of energy, with no losses to the surroundings).
Q12 — Principle of Calorimetry & Mixtures · easy · theory
The principle of calorimetry is a direct consequence of the law of:
A. Conservation of momentum
B. Conservation of energy ✓ Correct
C. Conservation of mass
D. Inertia
Solution: Heat lost = heat gained is simply energy conservation applied to thermal exchange.
Q13 — Principle of Calorimetry & Mixtures · easy · theory
When two liquids of EQUAL mass and EQUAL specific heat at temperatures T₁ and T₂ are mixed (no losses), the final temperature is:
A. Closer to the hotter one
B. √(T₁T₂)
C. (T₁ + T₂)/2 ✓ Correct
D. T₁ + T₂
Solution: Equal mc on both sides gives the simple mean. Speed-trick: any asymmetry (mass or c) pulls T_mix towards the larger mc side.
Q14 — Principle of Calorimetry & Mixtures · easy · theory
The final temperature of a mixture always lies:
A. Between the initial temperatures of the two bodies ✓ Correct
B. Below both initial temperatures
C. Above both initial temperatures
D. Exactly at the mean always
Solution: Heat flows from hot to cold until a common intermediate temperature is reached; T_mix can never lie outside the range [T_cold, T_hot].
Q15 — Principle of Calorimetry & Mixtures · easy · numerical
100 g of water at 80°C is mixed with 100 g of water at 20°C. The final temperature is:
A. 60°C
B. 40°C
C. 50°C ✓ Correct
D. 100°C
Solution: Equal masses of the same liquid: T_mix = (80 + 20)/2 = 50°C.
Q16 — Principle of Calorimetry & Mixtures · easy · numerical
200 g of water at 90°C is mixed with 100 g of water at 30°C. The final temperature is:
A. 50°C
B. 70°C ✓ Correct
C. 80°C
D. 60°C
Solution: T = (200×90 + 100×30)/(200 + 100) = (18000 + 3000)/300 = 70°C.
Q17 — Principle of Calorimetry & Mixtures · easy · numerical
50 g of water at 100°C is added to 200 g of water at 25°C. The final temperature is:
A. 35°C
B. 50°C
C. 40°C ✓ Correct
D. 62.5°C
Solution: T = (50×100 + 200×25)/250 = (5000 + 5000)/250 = 40°C.
Q18 — Principle of Calorimetry & Mixtures · easy · numerical
A 100 g copper calorimeter (c = 0.1 cal/g°C) has a water equivalent of:
A. 10 g ✓ Correct
B. 1 g
C. 1000 g
D. 100 g
Solution: W = mc/c_w = 100 × 0.1/1 = 10 g of water.
Q19 — Latent Heat & Phase Transitions · easy · theory
The latent heat of a substance is the heat required to:
A. Change the phase of unit mass WITHOUT any change in temperature ✓ Correct
B. Raise the temperature of unit mass by 1°C
C. Break the substance into atoms
D. Raise the temperature of one mole by 1°C
Solution: Q = mL: latent ("hidden") heat changes the phase at constant temperature — it goes into breaking intermolecular bonds, not raising temperature.
Q20 — Latent Heat & Phase Transitions · easy · theory
During melting or boiling, the temperature of a pure substance:
A. Falls
B. Fluctuates randomly
C. Rises steadily
D. Remains constant even though heat is being supplied ✓ Correct
Solution: All supplied heat goes into the phase change (latent heat), so the temperature stays constant until the change is complete — the flat portions of a heating curve.
Q21 — Latent Heat & Phase Transitions · easy · theory
The latent heat of vaporization of water compared to its latent heat of fusion is:
A. About half
B. Much smaller
C. Equal
D. Much larger (540 vs 80 cal/g) ✓ Correct
Solution: L_v = 540 cal/g vs L_f = 80 cal/g: separating molecules completely into vapour needs far more energy than loosening the solid lattice.
Q22 — Latent Heat & Phase Transitions · easy · theory
The boiling point of water, with an increase in pressure:
A. Decreases
B. Increases (principle of the pressure cooker) ✓ Correct
C. Becomes 0°C
D. Is unchanged
Solution: Higher pressure makes it harder for vapour to form, raising the boiling point — a pressure cooker cooks faster at ~120°C. At altitude, lower pressure lowers the boiling point.
Q23 — Latent Heat & Phase Transitions · easy · theory
Sublimation is the direct change from:
A. Liquid to solid
B. Solid to vapour without becoming liquid ✓ Correct
C. Vapour to liquid
D. Solid to liquid
Solution: Substances like camphor, dry ice (solid CO₂) and iodine pass directly from solid to vapour — sublimation.
Q24 — Latent Heat & Phase Transitions · easy · numerical
The heat required to melt 50 g of ice at 0°C (L_f = 80 cal/g) is:
A. 4000 cal ✓ Correct
B. 80 cal
C. 400 cal
D. 8000 cal
Solution: Q = mL_f = 50 × 80 = 4000 cal.
Q25 — Latent Heat & Phase Transitions · easy · numerical
The heat required to vaporize 20 g of water at 100°C (L_v = 540 cal/g) is:
A. 10800 cal ✓ Correct
B. 5400 cal
C. 1080 cal
D. 540 cal
Solution: Q = mL_v = 20 × 540 = 10800 cal.
Q26 — Latent Heat & Phase Transitions · easy · numerical
The heat needed to convert 10 g of ice at 0°C into water at 20°C is:
A. 1000 cal ✓ Correct
B. 200 cal
C. 1200 cal
D. 800 cal
Solution: Melt: 10 × 80 = 800 cal; warm: 10 × 1 × 20 = 200 cal. Total = 1000 cal.
Q27 — Latent Heat & Phase Transitions · easy · numerical
The heat required to convert 5 g of water at 100°C completely into steam at 100°C is:
A. 2900 cal
B. 540 cal
C. 500 cal
D. 2700 cal ✓ Correct
Solution: Q = mL_v = 5 × 540 = 2700 cal (temperature unchanged — pure phase change).
Q28 — Latent Heat & Phase Transitions · easy · numerical
The heat released when 15 g of steam at 100°C condenses into water at 100°C is:
A. 1200 cal
B. 8100 cal ✓ Correct
C. 4050 cal
D. 540 cal
Solution: Condensation releases the same latent heat: Q = 15 × 540 = 8100 cal.
Q29 — Ice-Water-Steam Mixing Problems · easy · theory
The heat-flow sequence for converting ice at −10°C to steam at 100°C is:
A. Melt → warm ice → vaporize → warm water
B. Vaporize → melt → warm
C. A single continuous temperature rise
D. Warm ice → melt at 0°C → warm water → vaporize at 100°C ✓ Correct
Solution: Four stages: m·c_i·(10) + m·L_f + m·c_w·(100) + m·L_v — two warmings and two flat phase changes on the heating curve.
Q30 — Ice-Water-Steam Mixing Problems · easy · theory
On a temperature-vs-heat (heating) curve of ice being converted to steam, the flat (horizontal) portions represent:
A. Phase changes at constant temperature (melting and boiling) ✓ Correct
B. Cooling stages
C. Rapid temperature rise
D. Loss of mass
Solution: Flat segments = latent-heat stages (0°C melting, 100°C boiling); sloped segments = specific-heat warming. The boiling plateau is much longer (540 vs 80 cal/g).