Specific Heat Calorimetry

8 MCQs9-step worked example

Source: NCERT Properties of Bulk MatterPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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 11, page 5). 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.

Practice 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:

MCQ 2Easy RecallPractice

When heat is supplied to a body undergoing a phase change at constant pressure, its temperature:

MCQ 3Easy RecallPractice

In calorimetry, the principle of method of mixtures is based on:

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:

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:

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:

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. The equilibrium temperature is 30 °C. The initial temperature of the iron block is closest to:

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)

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. ---

Before solving, remember these

Formula

Heat capacity and specific heat

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, p. 5

Formulas

12 formulas — click to collapse

Bernoulli's equation

P + \frac{1}{2} ρ v^{2} + ρ g h = const

Conservation of energy along a streamline of incompressible non-viscous flow.

Symbol	Quantity	SI Unit
P	pressure	Pa
rho	density	kg/m^3
v	speed	m/s
g	gravity	m/s^2
h	height	m

Valid when

Steady, non-viscous, incompressible flow
Along a single streamline
No work added/removed

Bulk modulus

K = - V \frac{d P}{d V}

Resistance of a material to uniform compression. Inverse: compressibility.

Symbol	Quantity	SI Unit
K	bulk modulus	Pa
V	volume	m^3
P	pressure	Pa

Valid when

Isotropic compression
Within elastic regime

Capillary rise/depression

h = \frac{2 T c o s θ}{ρ g r}

Height a liquid rises (or falls) in a capillary tube. cos(theta) > 0: rises (wetting); < 0: depresses.

Symbol	Quantity	SI Unit
h	capillary height	m
T	surface tension	N/m
theta	contact angle	rad
rho	density	kg/m^3
r	tube radius	m

Valid when

Narrow tube (capillary regime)
Constant theta

Pressure in static fluid

P = P_{0} + ρ g h

Pressure at depth h below free surface of fluid of density rho.

Symbol	Quantity	SI Unit
P	total pressure	Pa
P0	atmospheric/surface pressure	Pa
rho	density	kg/m^3
h	depth	m

Valid when

Static fluid (no flow)
Constant g
Constant rho (incompressible)

Latent heat

Q = m L

Heat absorbed/released during phase change at constant T. L_fusion or L_vaporisation.

Symbol	Quantity	SI Unit
Q	heat	J
m	mass	kg
L	latent heat	J/kg

Valid when

Phase transition (constant T during)
All mass m undergoes the transition

Specific heat / heat capacity

Q = m c Δ T

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

Stefan-Boltzmann radiation law

P = σ ε A T^{4}

Radiation power from a body. Black body epsilon=1; net to surroundings P = sigma*epsilon*A*(T^4 - T_s^4).

Symbol	Quantity	SI Unit
sigma	Stefan-Boltzmann = 5.67e-8	W/m^2/K^4
epsilon	emissivity (0-1)	-
A	surface area	m^2
T	absolute temp	K

Valid when

Body in radiative equilibrium
T in kelvins

Stokes' law (viscous drag on sphere)

F = 6 π η r v

Drag force on a sphere of radius r moving with velocity v through viscous fluid (low Reynolds number).

Symbol	Quantity	SI Unit
F	drag force	N
eta	viscosity	Pa*s
r	sphere radius	m
v	velocity	m/s

Valid when

Smooth, slow flow (low Reynolds number)
Spherical body
Newtonian fluid

Excess pressure inside drop/bubble

Δ P_{drop} = \frac{2 T}{r}; Δ P_{bubble} = \frac{4 T}{r}

Excess internal pressure due to surface tension. Bubble has 2 surfaces, hence factor 4.

Symbol	Quantity	SI Unit
Delta_P	excess pressure	Pa
T	surface tension	N/m
r	radius	m

Valid when

Spherical drop or bubble
Constant T (one fluid pair)

Terminal velocity of sphere in viscous fluid

v_{t} = \frac{2 r ^{2} g ( ρ _{s} - ρ _{f} )}{9 η}

Constant velocity reached when net force is zero (gravity balanced by buoyancy + viscous drag).

Symbol	Quantity	SI Unit
v_t	terminal velocity	m/s
r	sphere radius	m
rho_s	sphere density	kg/m^3
rho_f	fluid density	kg/m^3
eta	viscosity	Pa*s

Valid when

Steady state (net force zero)
Stokes regime applicable

Thermal expansion (linear/area/volume)

\frac{Δ L}{L} = α Δ T; \frac{Δ A}{A} = 2 α Δ T; \frac{Δ V}{V} = β Δ T

Fractional change in length, area, volume per degree temperature change.

Symbol	Quantity	SI Unit
alpha	linear coefficient	1/K
beta	volume coefficient	1/K
Delta_T	temperature change	K

Valid when

Isotropic material
Modest temperature range (alpha ~ constant)

Young's modulus

Y = \frac{F L}{A Δ L}

Ratio of longitudinal stress to longitudinal strain in a stretched wire/rod within elastic limit.

Symbol	Quantity	SI Unit
Y	Young's modulus	Pa
F	applied force	N
A	cross-section area	m^2
L	original length	m
Delta_L	extension	m

Valid when

Within elastic limit (Hooke's law region)
Uniform cross-section
Force along length

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

5 items — click to collapse

Trap

Soap bubble vs drop excess pressure

Category: Similar Terms

Student uses 2T/r for soap bubble (drop formula). Bubble has 2 surfaces → 4T/r.

When it triggers

Question mentions soap bubble OR liquid drop OR air bubble in liquid.

How to avoid

Drop in air: 1 surface → 2T/r. Soap bubble in air: 2 surfaces → 4T/r. Air bubble in liquid: 1 surface → 2T/r.

Trap

Young's vs bulk modulus confusion

Category: Similar Terms

Student uses Y formula when problem is about volumetric compression (use K) or vice versa.

When it triggers

Problem describes longitudinal stretching (use Y), volumetric pressure (use K), or shear (use G).

How to avoid

Y: longitudinal stress/strain. K: volumetric. G: shear. Match modulus to deformation type.

Mistake

Uses 2T/r excess-pressure formula for soap bubble. Bubble has 2 surfaces → 4T/r.

Root cause: concept gap

Correction

Drop in air: 1 surface → ΔP = 2T/r. Soap bubble: 2 surfaces (inner + outer) → ΔP = 4T/r. Air bubble inside liquid: 1 surface → 2T/r.

Mistake

Treats v_terminal as linear in r instead of r².

Root cause: formula misuse

Correction

v_t ∝ r² (because Stokes drag ∝ r v, gravity ∝ r³ - r³ = r³_diff). Doubling radius quadruples terminal velocity, not doubles.

Mistake

Uses Young's modulus formula for volumetric compression problem (or vice versa).

Root cause: formula misuse

Correction

Y for longitudinal stretch (FL/A·ΔL); K for volumetric compression (-V·dP/dV); G for shear. Match modulus type to deformation type before computing.

Past Year Questions

15 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse

NEET 2025

A balloon is made of a material of surface tension S and its inflation outlet (from where gas is filled in it) has small area A. It is filled with a gas of density ρ and takes a spherical shape of radius R. When the gas is allowed to flow freely out of it, its radius r changes from R to 0 (zero) in time T. If the speed v(r) of gas coming out of the balloon depends on r as ra and T ∝ Sα Aβ ργ Rδ then 1 1 1 1 7 1 1 3

1a= ,α= ,β=− ,γ= ,δ=

2a= ,α= ,β=−1,γ=+1,δ= 2 2 2 2 2 2 2 2 1 1 1 5 1 1 1 7

3a=− ,α=– ,β=−1,γ=− ,δ=

4a=− ,α=− ,β=−1,γ= ,δ= 2 2 2 2 2 2 2 2

NTA Answer: Option 4(final)

NEET 2025

Consider a water tank shown in the figure. It has one wall at x = L and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density ρ, the liquid surface makes angle θ 0 (θ 0 << 1) with the x-axis at x = L. If y(x) is the height of the surface then the equation for y(x) is: dy (takeθ(x)=sinθ(x)=tanθ(x)= ,g is the acceleration due to gravity) dx dy ρg d2y ρg

1= x

2= x dx S dx2 S d2y ρg d2y ρg

3= y

4= dx2 S dx2 S

NTA Answer: Option 3(final)

NEET 2024Revised key

The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young’s modulus, respectively, are 8 × 108 N m–2 and 2 × 1011 N m–2, is:

14 mm

20.4 mm

340 mm

48 mm

NTA Answer: Option 1(revised_final)

NEET 2024Revised key

A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 Nm–1, then the excess force required to take it away from the surface is

119.8 mN

2198 N

31.98 mN

499 N

NTA Answer: Option 1(revised_final)

NEET 2024Revised key

A metallic bar of Young’s modulus, 0.5 × 1011 N m–2 and coefficient of linear thermal expansion 10–5 °C–1, length 1 m and area of cross-section 10–3 m2 is heated from 0°C to 100°C without expansion or bending. The compressive force developed in it is :

15 × 103 N

250 × 103 N

3100 × 103 N

42 × 103 N

NTA Answer: Option 2(revised_final)

NEET 2023

The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly (surface tension of soap solution = 0.03 N m–1)

15.06 × 10–4 J

23.01 × 10–4 J

350.1 × 10–4 J

430.16 × 10–4 J

NTA Answer: Option 2(final)

NEET 2023

The venturi-meter works on

1Bernoulli’s principle

2The principle of parallel axes

3The principle of perpendicular axes

4Huygen’s principle

NTA Answer: Option 1(final)

NEET 2023

Let a wire be suspended from the ceiling (rigid support) and stretched by a weight W attached at its free end. The longitudinal stress at any point of cross-sectional area A of the wire is

1W/A

2W/2A

3Zero

42W/A

NTA Answer: Option 1(final)

NEET 2022

If a soap bubble expands, the pressure inside the bubble

1Is equal to the atmospheric pressure

2Decreases

3Increases

4Remains the same

NTA Answer: Option 2(final)

NEET 2022

A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball (v) as a function of time (t) is

NTA Answer: Option 3(final)

NEET 2022

Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The stretching of a spring is determined by the shear modulus of the material of the spring. Reason (R): A coil spring of copper has more tensile strength than a steel spring of same dimensions. In the light of the above statements, choose the most appropriate answer from the options given below

1(A) is false but (R) is true

2Both (A) and (R) are true and (R) is the correct explanation of (A)

3Both (A) and (R) are true and (R) is not the correct explanation of (A)

4(A) is true but (R) is false

NTA Answer: Option 4(final)

NEET 2021

The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of d glycerine is , then the viscous force acting on the 2 ball will be Mg

12Mg

22 3 Mg

3Mg

NTA Answer: Option 2(final)

NEET 2021

A cup coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is 5 13 t t

1(2) 13 10 13 10 t t

2d hc   2m c  λ2 λ =  

3(4) 5 13

4d h  

NTA Answer: Option 3(final)

NEET 2020

r U U 17. A capillary tube of radius r is immersed in water ° h – ÷ U and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary 5 g – 2r tube of radius 2r is immersed in water. The mass U – U — of water that will rise in this tube is :

22.5 g

35.0 g

NTA Answer: Option 4(final)

NEET 2020

r U r (r =1.5 r ) U U 45. The quantities of heat required to raise the 1 2 1 2 1 K fl h U fl c temperature of two solid copper spheres of radii — r and r (r =1.5 r ) through 1 K are in the 1 2 1 2 5 ratio :

127 3

29 8

33 4

NTA Answer: Option 2(final)

How NEET usually asks this

5 recurring patterns from past papers — click to collapse

Apply Bernoulli's equation to flow through pipes of varying cross-section / heights / Venturi-like geometries.

InterpretationMedium

Common distractors

forgets height term

Drops rho*g*h term

equates pressures incorrectly

Picks wrong reference points

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

Energy to form a bubble of given radius from soap solution; or pressure inside vs outside. Bubble has 2 surfaces (factor 4T/r).

Direct ApplicationEasy

Common distractors

uses 2T r instead of 4T r for bubble

Treats soap bubble like a drop

Sphere falling in viscous liquid; find terminal velocity or shape of v(t) curve. v_t ∝ r^2.

InterpretationMedium

Common distractors

expects linear acceleration

Default to constant acceleration without recognising drag-induced terminal v

Wire stretching: given Y, length, area, force or stress, find elongation or stress at limit. Y = FL/(A delta_L).

Multi StepMedium

Common distractors

uses bulk modulus formula

Confuses Y with K

Sources

NCERT refs: Class 11 Physics Chapter 11, p.5

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