Q = m c ΔT, where m is mass, c is specific heat capacity (J/kg/K). Heat capacity C = m c (J/K). For water: c = 4186 J/kg/K (one of the highest specific heats).
-- NCERT Class 11 Physics, Ch. 10, p. 208Specific Heat Calorimetry
8 MCQs6 revision cards9-step worked example
Source: NCERT Properties of Solids and LiquidsPYQ coverage: NEET 2020Official key: NTA-verifiedLast updated: 25 Sep 2026
Specific Heat Calorimetry, explained for NEET
The high-frequency trap in calorimetry problems is forgetting that heat exchange depends on mass × specific heat × temperature change — not on any single factor alone. Two bodies made of the same material but with different masses need different amounts of heat for the same temperature rise, and NEET distractors routinely isolate one variable to bait you.
Specific heat capacity (c) is the heat required to raise 1 kg of a substance by 1 K. The governing equation is Q = mcΔT, where Q is the heat transferred, m is the mass, and ΔT is the temperature change (NCERT Class 11 Physics Chapter 10, page 206). The SI unit of specific heat is J kg⁻¹ K⁻¹. Water has an unusually high specific heat (~4186 J kg⁻¹ K⁻¹), which is why it is the standard calorimetric liquid.
Calorimetry applies conservation of energy to heat exchange. When a hot body is placed in contact with a cold body inside an insulated calorimeter:
Heat lost by hot body = Heat gained by cold body
m₁c₁(T₁ − T_eq) = m₂c₂(T_eq − T₂)
where T_eq is the final equilibrium temperature. This equation assumes no phase change occurs — if one body reaches its melting or boiling point during the exchange, the latent heat formula Q = mL must be applied for the phase-change portion before continuing with Q = mcΔT for any further temperature change.
Watch out: When the problem gives two objects of the same material but different dimensions, the distractor that compares only specific heats (ignoring mass differences) is a common wrong option. Always compute or compare Q = mcΔT as a product — never drop a factor.
Can you answer these Specific Heat Calorimetry MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The SI unit of specific heat capacity is:
Show answer and why every option is right or wrong
Answer: D. Specific heat capacity is defined as heat per unit mass per unit temperature change. In SI, heat is in joules, mass in kilograms, and temperature in kelvins, giving J kg⁻¹ K⁻¹ (NCERT Class 11 Physics Chapter 10, page 206).
Why A is wrong: A is wrong because J mol⁻¹ K⁻¹ is the unit of molar specific heat (molar heat capacity), not specific heat capacity which is defined per unit mass.
Why B is wrong: B is wrong because W m⁻¹ K⁻¹ is the SI unit of thermal conductivity, not specific heat capacity.
Why C is wrong: C is wrong because cal g⁻¹ °C⁻¹ is the CGS unit, not the SI unit. NEET uses SI units exclusively.
MCQ 2Easy RecallPractice
When heat is supplied to a body undergoing a phase change at constant pressure, its temperature:
Show answer and why every option is right or wrong
Answer: A. During a phase change (melting or boiling), the supplied heat is used to overcome intermolecular forces (latent heat Q = mL), and temperature remains constant until the entire mass has changed phase (NCERT Class 11 Physics Chapter 10, page 208).
Why B is wrong: B is wrong because a linear temperature increase with heat applies only when there is no phase change (Q = mcΔT region). During a phase transition, temperature is constant despite continuous heat input.
Why C is wrong: C is wrong because temperature does not decrease when heat is being added. A decrease would require net heat loss from the body.
Why D is wrong: D is wrong because there is no temperature decrease during heating of a single body. This pattern does not describe any physical process during simple heat addition.
MCQ 3Easy RecallPractice
In calorimetry, the principle of method of mixtures is based on:
Show answer and why every option is right or wrong
Answer: B. The method of mixtures directly applies conservation of energy: in an insulated system, the heat lost by the hotter body equals the heat gained by the cooler body (NCERT Class 11 Physics Chapter 10, page 206).
Why A is wrong: A is wrong because conservation of momentum governs collisions and force interactions, not heat exchange between bodies.
Why C is wrong: C is wrong because Newton's law of cooling describes the rate of cooling of a body to its surroundings (dT/dt ∝ ΔT), not the equilibrium condition in calorimetry.
Why D is wrong: D is wrong because the zeroth law defines thermal equilibrium and the concept of temperature, but the quantitative heat-balance equation (heat lost = heat gained) comes from conservation of energy, not from the zeroth law.
MCQ 4Direct ApplicationPractice
A copper block of mass 2.0 kg at 100 °C is placed in 1.0 kg of water at 20 °C in an insulated calorimeter. Given c_copper = 390 J kg⁻¹ K⁻¹ and c_water = 4200 J kg⁻¹ K⁻¹, the equilibrium temperature is closest to:
Show answer and why every option is right or wrong
Answer: A. Heat lost by copper = heat gained by water: 2.0 × 390 × (100 − T) = 1.0 × 4200 × (T − 20). Solving: 780(100 − T) = 4200(T − 20) → 78000 − 780T = 4200T − 84000 → 162000 = 4980T → T ≈ 32.5 °C, closest to 35 °C.
Why B is wrong: B (25 °C) is wrong because it underestimates the copper's contribution. With 780 J/K heat capacity from the copper block, the equilibrium temperature is pushed higher than 25 °C.
Why C is wrong: C (45 °C) is wrong because it overestimates the copper's effect. Water has a much larger heat capacity (4200 J/K) compared to the copper block (780 J/K), so the final temperature stays closer to the water's initial temperature.
Why D is wrong: D (55 °C) is wrong because it ignores the large heat capacity of water. The water absorbs heat with minimal temperature rise, pulling the equilibrium well below the midpoint of 100 °C and 20 °C.
MCQ 5Direct ApplicationPractice
Two iron spheres A and B have radii r and 2r respectively. Both are heated through the same temperature rise ΔT. The ratio of heat absorbed Q_A : Q_B is:
Show answer and why every option is right or wrong
Answer: B. Since both spheres are iron, they share the same specific heat c and density ρ. Mass is proportional to volume: m ∝ r³. So Q = mcΔT ∝ r³. The ratio is r³ : (2r)³ = 1 : 8. This directly applies the pattern where ignoring mass difference leads to the wrong answer.
Why A is wrong: A (1 : 2) is wrong because this ratio compares radii, not masses. Heat depends on mass (∝ r³), not on radius directly.
Why C is wrong: C (1 : 4) is wrong because this ratio compares surface areas (∝ r²). Q = mcΔT depends on mass which is proportional to volume (r³), not surface area.
Why D is wrong: D (1 : 6) is wrong because there is no physical basis for a factor of 6 here. Volume ratio of spheres with radii r and 2r is 1 : 8, not 1 : 6.
MCQ 6Direct ApplicationPractice
10 g of ice at 0 °C is mixed with 10 g of water at 100 °C in an insulated container. Given L_fusion = 3.36 × 10⁵ J/kg and c_water = 4200 J kg⁻¹ K⁻¹, the final temperature of the mixture is:
Show answer and why every option is right or wrong
Answer: C. Heat available from the hot water cooling to 0 °C = 0.010 × 4200 × 100 = 4200 J. Heat needed to melt all the ice = 0.010 × 3.36 × 10⁵ = 3360 J, so all the ice melts and the final temperature is above 0 °C. Balancing heat lost and heat gained: 0.010 × 4200 × (100 − T) = 3360 + 0.010 × 4200 × T → 4200 − 42T = 3360 + 42T → 840 = 84T → T = 10 °C. The spare 840 J warms all 20 g of water, not just the 10 g of melt water.
Why A is wrong: A (0 °C) is wrong because the heat available from the hot water (4200 J) exceeds the heat needed to melt the ice (3360 J). The remaining 840 J raises the temperature above 0 °C.
Why B is wrong: B (40 °C) is wrong because this ignores the latent heat of fusion entirely. Without accounting for the 3360 J consumed during melting, the temperature is overestimated.
Why D is wrong: D (50 °C) is wrong because this assumes equal masses at equal specific heats without any phase change — a simple midpoint average. The latent heat of fusion consumes a large portion of the available energy, pulling the equilibrium temperature well below 50 °C.
MCQ 7CalculationPractice
A 500 g aluminium vessel (c_Al = 900 J kg⁻¹ K⁻¹) contains 300 g of water at 25 °C. A 200 g iron block (c_Fe = 450 J kg⁻¹ K⁻¹) heated to T_i is dropped in, and the mixture settles at 30 °C. Taking c_water = 4200 J kg⁻¹ K⁻¹ and neglecting losses, the initial temperature of the iron block is:
Show answer and why every option is right or wrong
Answer: D. D is correct. Heat gained by the water and the vessel, both rising 5 K from 25 °C to 30 °C: (0.300)(4200)(5) + (0.500)(900)(5) = 6300 + 2250 = 8550 J. The iron falls from T_i to 30 °C and must supply exactly that: (0.200)(450)(T_i − 30) = 90(T_i − 30) = 8550, so T_i − 30 = 95 and T_i = 125 °C.
Why A is wrong: A is wrong because 95 °C is T_i − 30, the temperature DROP of the iron, reported as its starting temperature. The last step adds back the 30 °C it fell to.
Why B is wrong: B is wrong because 100 °C comes from leaving out the aluminium vessel: 90(T_i − 30) = 6300 gives T_i = 100 °C. The vessel also warms by 5 K and takes 2250 J of the iron's heat.
Why C is wrong: C is wrong because 120 °C comes from using 25 °C as the iron's final temperature: 90(T_i − 25) = 8550. The iron ends at the equilibrium temperature, 30 °C, like everything else in the vessel.
MCQ 8Concept TrapPractice
Equal masses of water, copper, and aluminium are given the same amount of heat Q. Which statement correctly describes the final temperature ranking? (c_water > c_Al > c_Cu)
Show answer and why every option is right or wrong
Answer: C. From Q = mcΔT, for fixed m and Q: ΔT = Q/(mc). The substance with the smallest specific heat has the largest temperature rise. Since c_Cu < c_Al < c_water, the temperature rise follows ΔT_Cu > ΔT_Al > ΔT_water, giving T_Cu > T_Al > T_water (all starting from the same initial temperature).
Why A is wrong: A is wrong because it reverses the relationship. Higher specific heat means smaller temperature rise for the same heat input, not larger. Water, with the highest c, shows the smallest temperature increase.
Why B is wrong: B is wrong because it swaps the positions of copper and aluminium. Since c_Cu < c_Al, copper has a larger temperature rise than aluminium, not smaller.
Why D is wrong: D is wrong because equal heat input to equal masses does NOT produce the same temperature change when specific heats differ. ΔT is inversely proportional to c.
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Specific Heat Calorimetry: quick recall before you leave
How do you solve a Specific Heat Calorimetry question? A worked example
Pattern: Two bodies of given material and dimensions; find ratio of heat needed for the same temperature change (NEET pattern: heat capacity two bodies).
- 1
Given
Two solid cubes are made of the same metal (density ρ, specific heat c). Cube A has side length a = 0.10 m. Cube B has side length b = 0.20 m. Both are heated through the same temperature rise ΔT = 50 K.
- 2
Required
Find the ratio Q_A : Q_B.
- 3
Concept
Heat absorbed depends on mass, specific heat, and temperature change: Q = mcΔT. Since both cubes share the same material (same c and ρ) and the same ΔT, the ratio reduces to a comparison of masses. Mass = ρ × volume = ρ × (side)³.
- 4
Formula
Q = mcΔT, where m = ρ × side³.
- 5
Substitution
Q_A = ρ × (0.10)³ × c × 50
Q_B = ρ × (0.20)³ × c × 50
Q_A/Q_B = (0.10)³ / (0.20)³ - 6
Calculation
(0.10)³ = 1.0 × 10⁻³ m³
(0.20)³ = 8.0 × 10⁻³ m³
Q_A/Q_B = (1.0 × 10⁻³) / (8.0 × 10⁻³) = 1/8
Note on exact values: the side lengths 0.10 m and 0.20 m, the temperature rise 50 K, and the density and specific heat are treated as exact problem-defined values. They do not limit significant figures in the ratio. - 7
Final answer
Q_A : Q_B = 1 : 8
Cube B requires 8 times more heat than cube A for the same temperature rise, because its volume (and hence mass) is 8 times larger. - 8
Common trap
The distractor that gives 1 : 2 compares side lengths (linear dimensions). The distractor 1 : 4 compares surface areas (∝ side²). Heat depends on mass, which scales as volume (∝ side³). Always trace Q = mcΔT back to mass, and mass back to volume when density is the same.
- 9
Similar NEET-style question
Two aluminium cylinders P and Q have the same height but radii r and 3r respectively. If both are heated through the same temperature rise, find Q_P : Q_Q.
Answer: Mass ∝ πr²h, so Q_P/Q_Q = r²/(3r)² = 1/9. The ratio is 1 : 9.
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What to remember before solving Specific Heat Calorimetry questions
Which Specific Heat Calorimetry formulas do you need for NEET?
1 formula — click to collapse
Specific heat / heat capacity
Heat required to raise mass m by temperature Delta_T. Specific heat c is material property.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Q | heat | J |
| m | mass | kg |
| c | specific heat | J/kg/K |
| Delta_T | temp change | K |
Valid when
- No phase change during heating
- c approximately constant in temp range
More in Properties of Bulk Matter: 5 exam traps and mistakes · 11 formulas · 4 question patterns from its other lessons.
Specific Heat Calorimetry questions from past NEET papers
1 question from NEET 2020. Answers verified against NTA official keys. — click to collapse
NEET 2020
15/3
227/8
39/4
43/2
NTA Answer: Option 2(final)
All 17 past-paper questions from Properties of Bulk Matter →
How does NEET ask about Specific Heat Calorimetry?
1 recurring pattern from past papers — click to collapse
Two bodies of given material and dimensions; find ratio of heat needed for same temp change. Q ∝ m c.
Direct ApplicationEasy
Common distractors
ignores mass difference
Compares only specific heats, not masses
Sources
NCERT refs: Class 11 Physics Chapter 10, p.206
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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