Change of State And Calorimetry

Q: What is the principle of calorimetry?

The principle of calorimetry says that when bodies at different temperatures are mixed in an insulated container, the heat lost by the hotter bodies equals the heat gained by the colder bodies, until all reach a common temperature. It is the law of conservation of energy for heat.

Q: What is latent heat?

Latent heat is the heat absorbed or released per unit mass when a substance changes state at constant temperature, Q = mL. For water Lf ≈ 80 cal g⁻¹ and Lv ≈ 540 cal g⁻¹.

Q: Why does temperature remain constant during a change of state?

During melting or boiling the heat supplied is used to overcome the attraction between molecules, which raises their potential energy. Their average kinetic energy, and so the temperature, does not change until the whole substance has changed state.

Q: Why is the latent heat of vaporisation greater than the latent heat of fusion?

Melting only loosens the molecules, which stay close together. Vaporisation separates them completely and the vapour expands a lot against the atmosphere, so much more energy is needed per kilogram.

Q: What is the triple point of water?

The triple point of water is the single temperature and pressure, 273.16 K and about 0.006 atm, at which ice, liquid water and water vapour coexist in equilibrium. It is used as the fixed point of the Kelvin scale.

Q: What is regelation?

Regelation is the melting of ice under increased pressure and its refreezing when the pressure is removed. It happens because pressure lowers the melting point of ice. A loaded wire passing through an ice block without splitting it is the classic demonstration.

Q: How are calorimetry questions asked in NEET?

NEET asks direct mixture problems, usually ice with water or steam with water, the heat needed to turn ice into steam, and reading a heating curve. Always check whether all the ice melts before finding the final temperature.

Q: What calorimetry problems come in JEE Main?

JEE Main uses mixtures with phase changes, water equivalent of a calorimeter, heating curves at constant power, the ratio of latent heats from graph lengths, and energy conversion from falling or moving bodies into melting ice.

PhysicsThermal Properties Of MatterFor NEET aspirants

Calorimetry is the measurement of heat using one rule: in an isolated system, heat lost by hot bodies equals heat gained by cold bodies. Change of state adds latent heat, $Q = m L$ , absorbed or released at constant temperature. Together, change of state and calorimetry give the mixture problems (ice, water and steam) that appear in almost every NEET and JEE Main paper. This page covers the law of mixtures, latent heats, the heating curve, the effect of pressure, phase diagrams and a fail-safe method for ice-water problems.

On this page1Principle of calorimetry2Calorimeter3States and changes4Latent heat5Heating curve6Pressure effects7Phase diagrams8Mixture method

Key Formulas - Quick Reference

★ Must learnPrinciple of calorimetry: heat lost $=$ heat gained, $m_{1} s_{1} (T_{1} - T_{m}) = m_{2} s_{2} (T_{m} - T_{2})$
Mixture temperature (no phase change): $T_{m} = \frac{m _{1} s _{1} T _{1} + m _{2} s _{2} T _{2} + \dots}{m _{1} s _{1} + m _{2} s _{2} + \dots}$
★ Must learnLatent heat: $Q = m L$ , at constant temperature
★ Must learnIce: $L_{f} = 80 cal g^{- 1} \approx 3.36 \times 1 0^{5} J kg^{- 1}$ ; water: $L_{v} = 540 cal g^{- 1} \approx 2.26 \times 1 0^{6} J kg^{- 1}$
$s_{ice} \approx s_{steam} \approx 0.5 cal g^{- 1}^{\circ} C^{- 1}$ ; $s_{water} = 1 cal g^{- 1}^{\circ} C^{- 1}$
Heating curve at constant power $P$ : slope $\frac{d T}{d t} = \frac{P}{m s}$ ; flat time $t = \frac{m L}{P}$
Calorimeter of water equivalent $W$ : add $W$ to the mass of water in the heat balance
Triple point of water: $273.16 K$ , $0.006 atm$ ( $611 Pa$ )

1. Principle of Calorimetry (Law of Mixtures)

★ Must learn

When bodies at different temperatures are mixed in an insulated container, heat flows until they reach a common temperature $T_{m}$ . If no heat is lost to the surroundings, heat lost by the hot bodies = heat gained by the cold bodies. This is energy conservation applied to heat.

m_{1} s_{1} (T_{1} - T_{m}) = m_{2} s_{2} (T_{m} - T_{2}) (T_{1} > T_{m} > T_{2})

Written with signs, the total heat exchanged is zero: $\sum m_{i} s_{i} (T_{m} - T_{i}) = 0$ , which gives the weighted average

T_{m} = \frac{m _{1} s _{1} T _{1} + m _{2} s _{2} T _{2} + m _{3} s _{3} T _{3}}{m _{1} s _{1} + m _{2} s _{2} + m _{3} s _{3}}

valid when no body changes state. The "weights" are the heat capacities $m_{i} s_{i}$ , so $T_{m}$ lies nearer the temperature of the body with the larger heat capacity.

Figure 1: Principle of calorimetry (law of mixtures). Both bodies end at the same temperature

T_{m}

, which always lies between

T_{2}

and

T_{1}

; if no heat escapes, heat lost

=

heat gained.

1.1 The calorimeter

A calorimeter is a thin copper or aluminium can (small heat capacity, reaches the common temperature quickly) with a stirrer and a thermometer, kept inside an insulating jacket with a lid. The can itself takes part in the exchange: include its heat capacity $m_{c} s_{c}$ , or its water equivalent $W$ , on the side it belongs to.

Figure 2: A calorimeter: a thin copper can (small heat capacity, conducts quickly) with stirrer and thermometer, kept in an insulating jacket so that heat lost = heat gained holds inside it.

Exam Trick

Equal masses of the same liquid: $T_{m}$ is the plain average. For equal heat capacities, $T_{m} = \frac{T _{1} + T _{2}}{2}$ . For $n$ identical portions at different temperatures, $T_{m}$ is the mean of the temperatures. Always check that your $T_{m}$ lies between the extreme temperatures.

2. States of Matter and Change of State

Matter exists as solid, liquid or gas. A change from one state to another is a change of state (phase change). While it happens, the heat supplied breaks or loosens the bonds between molecules (increases their potential energy) instead of making them move faster, so the temperature stays constant until the change is complete.

Change	Direction	Temperature name	Heat
Melting (fusion)	solid → liquid	melting point	absorbed
Freezing (solidification)	liquid → solid	freezing point (same as melting point)	released
Vaporisation (boiling)	liquid → vapour	boiling point	absorbed
Condensation	vapour → liquid	same as boiling point	released
Sublimation	solid → vapour directly	sublimation point	absorbed
Deposition	vapour → solid directly		released

Figure 3: Changes of state. Red arrows absorb heat (melting, vaporisation, sublimation); blue arrows release the same latent heat (freezing, condensation, deposition). Temperature stays constant during each change.

Evaporation

At all temperatures, from the surface only, slow and silent. Faster with larger area, wind, higher temperature and lower humidity. Causes cooling (sweat, earthen pots).

Boiling

At one temperature for a given pressure (vapour pressure $=$ outside pressure), throughout the liquid with bubbles, fast. Temperature stays fixed while it lasts.

3. Latent Heat

★ Must learn

The latent heat $L$ of a substance is the heat absorbed or released per unit mass during a change of state at constant temperature:

Q = m L

SI unit $J kg^{- 1}$ . Latent heat of fusion $L_{f}$ : solid ↔ liquid at the melting point. Latent heat of vaporisation $L_{v}$ : liquid ↔ vapour at the boiling point. Both are measured at 1 atm unless stated.

For water, take $L_{f} = 80 cal g^{- 1}$ ( $3.33$ to $3.36 \times 1 0^{5} J kg^{- 1}$ ) and $L_{v} = 540 cal g^{- 1}$ ( $2.26 \times 1 0^{6} J kg^{- 1}$ ) unless the question gives other values. $L_{v} > L_{f}$ for every substance, because separating molecules completely (liquid to gas, with a large expansion against the atmosphere) needs far more energy than loosening them (solid to liquid).

Substance	Melting point (°C)	$L_{f}$ ( $kJ kg^{- 1}$ )	Boiling point (°C)	$L_{v}$ ( $kJ kg^{- 1}$ )
Water	0	333	100	2256
Ethanol	−114	104	78	854
Mercury	−39	11.8	357	272
Lead	328	24.5	1750	868
Gold	1063	64.5	2660	1580
Nitrogen	−210	25.5	−196	201
Oxygen	−219	13.8	−183	213

Steam burns are worse than boiling-water burns: $1 g$ of steam at $10 0^{\circ} C$ gives $540 cal$ on condensing before it even starts cooling, about nine times what $1 g$ of boiling water gives cooling to body temperature.
Ice at $0^{\circ} C$ cools better than water at $0^{\circ} C$ : each gram absorbs $80 cal$ while melting, without warming.
Specific heat "during" a phase change would be $\frac{Q}{m \times 0} = \infty$ : the heat goes into latent heat, not into temperature.

Key idea

During a change of state the temperature is constant; the heat

Q = m L

changes the molecular potential energy, not the kinetic energy.

4. The Heating Curve

Supply heat steadily to ice below $0^{\circ} C$ and plot temperature against heat supplied (or time, at constant power). The graph has sloping parts (one state warming) and flat parts (a change of state):

Ice warms: slope $\frac{Δ T}{Δ Q} = \frac{1}{m s _{ice}}$ .
Ice melts at $0^{\circ} C$ : flat, length $m L_{f}$ .
Water warms: slope $\frac{1}{m s _{water}}$ , half as steep as for ice, since $s_{water} = 2 s_{ice}$ .
Water boils at $10 0^{\circ} C$ : flat, length $m L_{v}$ , the longest part ( $L_{v} > L_{f}$ ).
Steam warms: slope $\frac{1}{m s _{steam}}$ , about as steep as for ice.

Figure 4: Heating curve drawn to scale for Solved Example 3. Flat parts are phase changes (

Q = m L

); sloping parts have slope

\frac{Δ T}{Δ Q} = \frac{1}{m s}

, so ice and steam (

s = 0.5

) climb twice as steeply as water (

s = 1

). Boiling alone takes

54

of the

73.5 kcal

Exam Trick

Read a heating curve like a v-t graph. At constant power $P$ : slope of a sloping part $= \frac{P}{m s}$ (steeper means smaller heat capacity), and length of a flat part $= \frac{m L}{P}$ . So $\frac{s _{1}}{s _{2}} = \frac{slope _{2}}{slope _{1}}$ and $\frac{L _{v}}{L _{f}} = \frac{length of boiling step}{length of melting step}$ .

Figure 5: Break every heating problem into stages:

m s Δ T

for each state and

m L

at each change of state, then add. The latent-heat steps (bold) usually dominate.

Quick Recall: tap to check

Why is the water part of the heating curve less steep than the ice part?

Slope

= \frac{1}{m s}

and water's specific heat is twice that of ice.

Which flat step of the heating curve of water is longer, and why?

Boiling:

L_{v} = 540 cal g^{- 1}

is much larger than

L_{f} = 80 cal g^{- 1}

At what temperature does water evaporate?

At all temperatures; boiling happens only at the boiling point.

5. Effect of Pressure: Boiling Point and Melting Point

5.1 Boiling point rises with pressure

A liquid boils when its saturated vapour pressure equals the external pressure. Raise the pressure and a higher temperature is needed; lower it and the liquid boils sooner.

Pressure cooker: about 2 atm inside, water boils near $12 0^{\circ} C$ , so food cooks faster.
Hill stations: lower air pressure, water boils below $10 0^{\circ} C$ , so cooking takes longer.
In a closed flask of hot water, pouring cold water on the flask condenses the vapour, lowers the pressure, and the water boils again.

Figure 6: A liquid boils when its vapour pressure equals the pressure above it (curve from the Antoine equation for water). Lower pressure on mountains lowers the boiling point; a pressure cooker raises it, so food cooks faster.

5.2 Melting point and regelation

Substances that expand on melting (wax, most solids) have their melting point raised by pressure. Substances that contract on melting (ice, cast iron, bismuth) have it lowered by pressure: ice melts at about $0.007 5^{\circ} C$ below $0^{\circ} C$ for each extra atmosphere.

Regelation is the melting of ice under pressure and its refreezing when the pressure is removed. A loaded wire passes through a block of ice without cutting it into two; pressing snow makes a snowball; ice skates glide on a thin film of water under the blade.

Figure 7: Regelation. Pressure lowers the melting point of ice (about

0.007 5^{\circ} C

per atm), so ice melts under the wire and refreezes above it. The same effect helps ice skates glide and snowballs stick.

6. Triple Point and Phase Diagrams

A phase diagram plots pressure against temperature and shows which state is stable. It has three curves:

Sublimation curve (solid ↔ vapour), fusion curve (solid ↔ liquid) and vaporisation curve (liquid ↔ vapour). On a curve two phases coexist; between curves one phase exists.
The three curves meet at the triple point, the unique temperature and pressure at which solid, liquid and vapour coexist in equilibrium. For water: $273.16 K$ ( $0.0 1^{\circ} C$ ) and $6.11 \times 1 0^{2} Pa$ ( $0.006 atm$ ). For $C O_{2}$ : $216.55 K$ ( $- 56. 6^{\circ} C$ ) and $5.11 atm$ .
The vaporisation curve ends at the critical point ( $37 4^{\circ} C$ , $218 atm$ for water); above it liquid and vapour cannot be told apart and $L_{v} = 0$ .

Figure 8: Phase diagrams (schematic, not to scale). Three curves meet at the triple point (TP), where solid, liquid and vapour coexist; the vaporisation curve ends at the critical point (C). Water's melting line slopes backwards (ice melts under pressure);

C O_{2}

's triple point lies above 1 atm, so dry ice sublimes.

Because the triple point of $C O_{2}$ is above 1 atm, liquid $C O_{2}$ cannot exist at atmospheric pressure: solid $C O_{2}$ ("dry ice") sublimes directly at $- 78. 5^{\circ} C$ . Water's fusion curve slopes to the left because ice is less dense than water.

JEE Advanced

The slope of a phase boundary is given by the Clausius-Clapeyron equation, $\frac{d p}{d T} = \frac{L}{T ( v _{2} - v _{1} )}$ , where $v_{1}$ , $v_{2}$ are specific volumes before and after the change. For ice to water, $v_{2} < v_{1}$ , so $\frac{d p}{d T} < 0$ : the melting point falls as pressure rises. For boiling, $v_{2} ≫ v_{1}$ and $\frac{d p}{d T} > 0$ .

7. Solving Mixture Problems with a Phase Change

When ice or steam is involved, the final state is not known in advance. Test at the phase-change temperature before writing the full balance:

Find the heat the hot side can give while cooling to $0^{\circ} C$ (or to $10 0^{\circ} C$ for steam problems).
Find the heat the cold side needs to reach $0^{\circ} C$ and change state completely.
Compare. If the hot side has more, the whole of the cold side changes state and $T_{m}$ is found from the full balance. If less, $T_{m}$ is the phase-change temperature and only part changes state: mass changed $= \frac{available heat}{L}$ .

Figure 9: The

0^{\circ} C

test. Never write one heat balance with an unknown

T_{m}

until you know whether all the ice melts; otherwise you may get an impossible answer such as

T_{m} < 0

with liquid water.

Exam Trick

Steam mixed with cold water: test at $10 0^{\circ} C$ . Heat to bring all the water to $10 0^{\circ} C$ versus heat all the steam gives on condensing ( $m L_{v}$ ) plus cooling to $10 0^{\circ} C$ . If the water needs less, $T_{m} = 10 0^{\circ} C$ and some steam remains. One gram of steam can heat about $6.4 g$ of water from $0$ to $10 0^{\circ} C$ ( $640 \div 100$ ).

Quick Recall: tap to check

10 g

ice at

0^{\circ} C

is mixed with

10 g

water at

4 0^{\circ} C

. Final state?

Water can give only

400 cal

(less than

800

T_{m} = 0^{\circ} C

5 g

ice melts.

What is the triple point of water?

273.16 K

and

0.006 atm

, where ice, water and vapour coexist.

Why does dry ice not melt at 1 atm?

Its triple point is at

5.11 atm

; below that there is no liquid phase, so it sublimes.

Figure 10: Mind map of this concept for quick revision.

8. Solved Examples

Solved Example 1

Equal masses of three liquids A, B and C are at

1 0^{\circ} C

1 5^{\circ} C

and

2 0^{\circ} C

. When A and B are mixed the temperature is

1 3^{\circ} C

; when B and C are mixed it is

1 6^{\circ} C

. What is the temperature when A and C are mixed?

Solution:

A and B: $m s_{1} (13 - 10) = m s_{2} (15 - 13) \Rightarrow 3 s_{1} = 2 s_{2}$ ...(1)

B and C: $m s_{2} (16 - 15) = m s_{3} (20 - 16) \Rightarrow s_{2} = 4 s_{3}$ ...(2)

From (1) and (2): $s_{1} = \frac{2}{3} s_{2} = \frac{8}{3} s_{3}$ .

A and C: $s_{1} (θ - 10) = s_{3} (20 - θ) \Rightarrow \frac{8}{3} (θ - 10) = 20 - θ \Rightarrow 11 θ = 140$ .

Answer: $θ = \frac{140}{11}^{\circ} C \approx 12. 7^{\circ} C$ .

Figure 11: Solved Example 1 on a number line. The mixture temperature is a weighted average

T_{m} = \frac{m _{1} s _{1} T _{1} + m _{2} s _{2} T _{2}}{m _{1} s _{1} + m _{2} s _{2}}

, so it sits closer to the liquid with the larger heat capacity.

Solved Example 2

Three liquids of masses

m_{1}

m_{2}

m_{3}

, specific heats

s_{1}

s_{2}

s_{3}

and temperatures

T_{1}

T_{2}

T_{3}

are mixed. Find the temperature of the mixture at thermal equilibrium (no change of state, no loss).

Solution:

The total heat exchanged is zero: $m_{1} s_{1} (T - T_{1}) + m_{2} s_{2} (T - T_{2}) + m_{3} s_{3} (T - T_{3}) = 0$ .

Answer:

T = \frac{m _{1} s _{1} T _{1} + m _{2} s _{2} T _{2} + m _{3} s _{3} T _{3}}{m _{1} s _{1} + m _{2} s _{2} + m _{3} s _{3}}

Writing every term as $m s (T - T_{i})$ avoids deciding in advance which liquids gain and which lose heat.

Solved Example 3

Find the heat needed to convert

100 g

of ice at

- 1 0^{\circ} C

into steam at

12 0^{\circ} C

. (

s_{ice} = s_{steam} = 0.5

s_{water} = 1 cal g^{- 1}^{\circ} C^{- 1}

L_{f} = 80

L_{v} = 540 cal g^{- 1}

)

Solution:

Ice $- 10 \to 0^{\circ} C$ : $100 \times 0.5 \times 10 = 500 cal$ .

Melting: $100 \times 80 = 8000 cal$ .

Water $0 \to 10 0^{\circ} C$ : $100 \times 1 \times 100 = 10000 cal$ .

Boiling: $100 \times 540 = 54000 cal$ .

Steam $100 \to 12 0^{\circ} C$ : $100 \times 0.5 \times 20 = 1000 cal$ .

Answer: $Q = 73500 cal = 73.5 kcal$ (heat absorbed; see Figures 4 and 5).

Solved Example 4

500 g

of water at

8 0^{\circ} C

is mixed with

100 g

of steam at

12 0^{\circ} C

. Find the final temperature and composition of the mixture. (

s_{steam} = 0.5 cal g^{- 1}^{\circ} C^{- 1}

L_{v} = 540 cal g^{- 1}

)

Solution:

Test at $10 0^{\circ} C$ . Heat the water needs to reach $10 0^{\circ} C$ : $500 \times 1 \times 20 = 10000 cal$ .

Heat the steam gives cooling to $10 0^{\circ} C$ : $100 \times 0.5 \times 20 = 1000 cal$ . The remaining $9000 cal$ must come from condensing steam; all of it could give $100 \times 540 = 54000 cal$ , far more than needed. So $T_{m} = 10 0^{\circ} C$ .

Steam condensed $= \frac{9000}{540} = 16.7 g$ .

Answer: $T_{m} = 10 0^{\circ} C$ with about $83.3 g$ of steam and $516.7 g$ of water.

Solved Example 5

100 g

of ice at

0^{\circ} C

is added to

200 g

of water at

3 0^{\circ} C

. Find the final temperature and the mass of ice left. (

L_{f} = 80 cal g^{- 1}

)

Solution:

Heat the water can give cooling to $0^{\circ} C$ : $200 \times 1 \times 30 = 6000 cal$ .

Heat to melt all the ice: $100 \times 80 = 8000 cal$ . Since $6000 < 8000$ , not all ice melts and $T_{m} = 0^{\circ} C$ .

Ice melted $= \frac{6000}{80} = 75 g$ .

Answer: $T_{m} = 0^{\circ} C$ ; $25 g$ of ice remains (with $275 g$ of water).

Solved Example 6

50 g

of ice at

- 1 0^{\circ} C

is put into

100 g

of water at

6 0^{\circ} C

. Find the final temperature. (

s_{ice} = 0.5 cal g^{- 1}^{\circ} C^{- 1}

L_{f} = 80 cal g^{- 1}

)

Solution:

Test: ice needs $50 \times 0.5 \times 10 + 50 \times 80 = 250 + 4000 = 4250 cal$ to become water at $0^{\circ} C$ ; the water can give $100 \times 60 = 6000 cal$ . So all the ice melts and $T_{m} > 0$ .

Balance: $100 (60 - T) = 4250 + 50 (T - 0) \Rightarrow 6000 - 4250 = 150 T$ .

Answer: $T = \frac{35}{3} \approx 11. 7^{\circ} C$ .

Solved Example 7

A sphere of aluminium of mass

0.047 kg

10 0^{\circ} C

is put into a copper calorimeter of mass

0.14 kg

containing

0.25 kg

of water at

2 0^{\circ} C

. The final temperature is

2 3^{\circ} C

. Find the specific heat of aluminium. (

s_{Cu} = 386 J kg^{- 1} K^{- 1}

s_{w} = 4180 J kg^{- 1} K^{- 1}

)

Solution:

Heat gained by water and calorimeter: $(0.25 \times 4180 + 0.14 \times 386) (23 - 20) = (1045 + 54.0) (3) = 3297 J$ .

Heat lost by aluminium: $0.047 \times s \times (100 - 23) = 3.619 s$ .

Answer: $s = \frac{3297}{3.619} \approx 911 J kg^{- 1} K^{- 1}$ .

Solved Example 8

Calculate the heat needed to convert

3 kg

of ice at

- 1 2^{\circ} C

into steam at

10 0^{\circ} C

at 1 atm. (

s_{ice} = 2100

s_{w} = 4186 J kg^{- 1} K^{- 1}

L_{f} = 3.35 \times 1 0^{5}

L_{v} = 2.256 \times 1 0^{6} J kg^{- 1}

)

Solution:

$Q_{1} = 3 \times 2100 \times 12 = 7.56 \times 1 0^{4} J$ ; $Q_{2} = 3 \times 3.35 \times 1 0^{5} = 1.005 \times 1 0^{6} J$ .

$Q_{3} = 3 \times 4186 \times 100 = 1.256 \times 1 0^{6} J$ ; $Q_{4} = 3 \times 2.256 \times 1 0^{6} = 6.768 \times 1 0^{6} J$ .

Answer: $Q = 9.10 \times 1 0^{6} J$ .

Solved Example 9

100 g

metal piece at

10 0^{\circ} C

is dropped into

180 g

of water at

2 0^{\circ} C

in a calorimeter of water equivalent

20 g

. The final temperature is

2 5^{\circ} C

. Find the specific heat of the metal.

Solution:

Heat gained by water and calorimeter: $(180 + 20) (1) (25 - 20) = 1000 cal$ .

Heat lost by the metal: $100 \times s \times 75 = 7500 s$ .

Answer: $s = \frac{1000}{7500} \approx 0.133 cal g^{- 1}^{\circ} C^{- 1}$ .

Solved Example 10

What mass of steam at

10 0^{\circ} C

must be passed into

1 kg

of water at

2 0^{\circ} C

to raise its temperature to

8 0^{\circ} C

? (

L_{v} = 540 cal g^{- 1}

)

Solution:

Each gram of steam gives $540$ (condensing) $+ 20$ (cooling from $100$ to $8 0^{\circ} C$ ) $= 560 cal$ .

Water needs $1000 \times 1 \times 60 = 60000 cal$ . So $560 m = 60000$ .

Answer: $m \approx 107 g$ .

Solved Example 11

Ice at

0^{\circ} C

is heated at a constant rate and melts completely in

8 min

. How much more time does the water take to reach

10 0^{\circ} C

? (

L_{f} = 80 cal g^{- 1}

)
(A) 8 min
(B) 10 min
(C) 12.5 min
(D) 54 min

Solution:

Answer: (B). At constant power, time $\propto$ heat: $\frac{t}{8} = \frac{m ( 1 ) ( 100 )}{m ( 80 )}$ , so $t = 10 min$ .

Solved Example 12

Why does

1 g

of steam at

10 0^{\circ} C

burn more than

1 g

of water at

10 0^{\circ} C

? Compare the heat each gives to skin at

3 7^{\circ} C

Solution:

Steam: $540$ (condensing) $+ 1 \times 63$ (cooling to $3 7^{\circ} C$ ) $= 603 cal$ . Water: $63 cal$ only.

Answer: steam gives about $603 cal$ , nearly ten times the $63 cal$ from boiling water, because of its latent heat.

Practice Questions

$10 g$ of ice at $0^{\circ} C$ is mixed with $10 g$ of water at $5 0^{\circ} C$ . Find the final temperature and the ice melted.Answer: $0^{\circ} C$ ; $6.25 g$
$20 g$ of steam at $10 0^{\circ} C$ is passed into $200 g$ of water at $2 0^{\circ} C$ (negligible container). Find the final temperature.Answer: about $76. 4^{\circ} C$
How much heat (in kJ) turns $1 kg$ of ice at $0^{\circ} C$ into water at $5 0^{\circ} C$ ? ( $L_{f} = 336 kJ kg^{- 1}$ , $s_{w} = 4.2 kJ kg^{- 1} K^{- 1}$ )Answer: $546 kJ$
A $200 g$ metal piece at $15 0^{\circ} C$ is dropped into $150 g$ of water at $2 7^{\circ} C$ ; the final temperature is $4 0^{\circ} C$ . Find the metal's specific heat.Answer: about $0.089 cal g^{- 1}^{\circ} C^{- 1}$
Why does the temperature stay constant while ice melts, although heat is supplied?Answer: The heat increases the potential energy of the molecules (breaks the lattice), not their kinetic energy
Why does food cook faster in a pressure cooker?Answer: Higher pressure raises the boiling point to about $12 0^{\circ} C$
In the heating curve at constant power, the melting step lasts $2 min$ and the boiling step $13.5 min$ . Find $\frac{L _{v}}{L _{f}}$ .Answer: $6.75$

Common Mistakes to Avoid

Watch out

Writing one heat balance with unknown $T_{m}$ for ice + water without the $0^{\circ} C$ test, and getting $T_{m} < 0^{\circ} C$ with liquid water left.
Using $s = 1$ for ice or steam. Take $s_{ice} \approx s_{steam} \approx 0.5 cal g^{- 1}^{\circ} C^{- 1}$ .
Forgetting the calorimeter (or its water equivalent) on the heat-gaining side.
Adding latent heat when there is no change of state, or forgetting it when there is one.
Assuming steam condenses completely. Test at $10 0^{\circ} C$ first: often only part of it condenses.
Mixing units: calories with kilograms or joules with grams. With $L = 80$ and $s = 1$ , masses must be in grams and heat in calories.
Saying pressure always raises the melting point. It lowers the melting point of ice (it contracts on melting).
Confusing evaporation (any temperature, surface) with boiling (one temperature, throughout the liquid).

Frequently Asked Questions

What is the principle of calorimetry?

The principle of calorimetry says that when bodies at different temperatures are mixed in an insulated container, the heat lost by the hotter bodies equals the heat gained by the colder bodies, until all reach a common temperature. It is the law of conservation of energy for heat.

What is latent heat?

Latent heat is the heat absorbed or released per unit mass when a substance changes state at constant temperature, $Q = m L$ . For water $L_{f} \approx 80 cal g^{- 1}$ and $L_{v} \approx 540 cal g^{- 1}$ .

Why does temperature remain constant during a change of state?

During melting or boiling the heat supplied is used to overcome the attraction between molecules, which raises their potential energy. Their average kinetic energy, and so the temperature, does not change until the whole substance has changed state.

Why is the latent heat of vaporisation greater than the latent heat of fusion?

Melting only loosens the molecules, which stay close together. Vaporisation separates them completely and the vapour expands a lot against the atmosphere, so much more energy is needed per kilogram.

What is the triple point of water?

The triple point of water is the single temperature and pressure, $273.16 K$ and about $0.006 atm$ , at which ice, liquid water and water vapour coexist in equilibrium. It is used as the fixed point of the Kelvin scale.

What is regelation?

Regelation is the melting of ice under increased pressure and its refreezing when the pressure is removed. It happens because pressure lowers the melting point of ice. A loaded wire passing through an ice block without splitting it is the classic demonstration.

How are calorimetry questions asked in NEET?

NEET asks direct mixture problems, usually ice with water or steam with water, the heat needed to turn ice into steam, and reading a heating curve. Always check whether all the ice melts before finding the final temperature.

What calorimetry problems come in JEE Main?

JEE Main uses mixtures with phase changes, water equivalent of a calorimeter, heating curves at constant power, the ratio of latent heats from graph lengths, and energy conversion from falling or moving bodies into melting ice.

Previous year questions on Change of State And Calorimetry

6 questions from past papers, each with a step-by-step solution.

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