JEE Mains Physics · Thermal Properties of Matter

Calorimetry and Latent Heat

Heat changes a body's temperature by Q = msΔT, or changes its phase at a fixed temperature by Q = mL; in a mixture with no losses, the heat lost by the hot parts equals the heat gained by the cold ones.

All Thermal Properties of Matter notes JEE Mains Physics strategy

Why this matters

Nineteen PYQs, five of them asking for a number, and four from 2026. Seven turn electrical, mechanical or chemical energy into heat, eight follow a substance through a change of phase or read a heating curve, and four mix ice and water or two liquids. Write every stage as its own term, and check whether all the ice melts before assuming it does.

Concept 1 of 3: Heat from a heater, a fuel or lost kinetic energy

Heat is energy, so any energy that disappears as work, motion or fuel can reappear as heat. Find how much energy goes into the body, multiply by the fraction that actually heats it, and set that equal to msΔT, plus mL if the body also melts. When the energy comes from the body's own motion or fall, its mass is on both sides and cancels.

Definition

$Q = ms\Delta T$ ; heat capacity $ms$ is in J/K, specific heat s in J kg⁻¹ K⁻¹.
Heater: useful power $= \eta P$ , so $\eta Pt = ms\Delta T$ .
Flowing water (geyser): heat per second $= \dot{m}s\Delta T$ ; fuel burnt per second $= \dot{m}s\Delta T \div$ heat of combustion.
Falling water: $mgh = ms\Delta T$ , so $\Delta T = gh/s$ , the same for any mass.
A moving body stopped: a fraction f of $\tfrac{1}{2}mv^{2}$ heats it; a bullet that melts needs $f \cdot \tfrac{1}{2}v^{2} = s\Delta T + L$ per kilogram.
1 cal = 4.2 J; water has $s = 1\ \text{cal g}^{-1}\ ^{\circ}C^{-1} = 4200\ \text{J kg}^{-1}\ \text{K}^{-1}$ .

Energy into heat

Q = ms\Delta T \qquad \eta Pt = ms\Delta T \qquad f\cdot\tfrac{1}{2}mv^{2} = ms\Delta T + mL

Worked example

An electric kettle of power 1500 W is 80% efficient. How long does it take to heat 1.5 kg of water from

20^{\circ}C

100^{\circ}C

? (

s = 4200\ \text{J kg}^{-1}\ \text{K}^{-1}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 23 January 2025 · Q12Moderate

Example 1 · Thermal Properties of Matter · Calorimetry and Latent Heat

A gun fires a lead bullet of temperature 300 K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J , then the mass of the bullet is

\_\_\_\_\_

grams. (Latent heat of fusion of lead

= 2.5 \times10^{4}JKg^{- 1}

and specific heat capacity of lead

= 125JKg^{- 1}{\text{ }K}^{- 1}

)

Grams and kilograms in one equation

Specific heats are given per gram or per kilogram, and heats of combustion per gram. Put every quantity in one system before dividing, or the answer is off by a thousand.

Stopping at the melting point

A body that heats up AND melts needs msΔT to reach its melting point and then mL more. Leaving out either term gives a mass or speed that is wrong by a large factor.

Which g the options use

Answers built on g = 9.8 and g = 10 differ by 2%, and both may appear as options. Use the value the question gives. If it gives none, work with 9.8 and check which option the result matches.

Drill 6 more on heat from a heater, a fuel or lost kinetic energy

Concept 2 of 3: Latent heat and the heating curve

While ice melts or water boils, the heat goes into breaking the bonds between molecules, not into raising the temperature. So a graph of temperature against heat supplied rises, goes flat at 0°C while the ice melts, rises again, and goes flat at 100°C while the water boils. The flat for boiling is much longer than the flat for melting, because water's latent heat of vaporisation is nearly seven times its latent heat of fusion.

Definition

Change of phase at constant temperature: $Q = mL$ . Latent heat L is in J/kg.
Ice at $-T_1$ to steam at $T_2 > 100^{\circ}C$ : five terms, $ms_{ice}T_1 + mL_f + ms_w(100) + mL_v + ms_{steam}(T_2 - 100)$ .
On a heating curve the slope of a rising part is $\dfrac{dT}{dQ} = \dfrac{1}{ms}$ : a smaller specific heat gives a steeper line.
With a constant heater, flat lengths are in the ratio of the latent heats: $L_v/L_f = 2256/336 \approx 6.7$ .
A curve stops where the process stops: ice heated to steam at 100°C ends on the boiling flat.
Two samples of equal mass on the same heater: the one whose temperature rises more slowly has the larger specific heat.
Units: heat capacity J/K, specific heat J kg⁻¹ K⁻¹, latent heat J/kg, thermal conductivity W m⁻¹ K⁻¹.

Latent heat and heating-curve slope

Q = mL \qquad \frac{dT}{dQ} = \frac{1}{ms}

Worked example

How much heat turns 0.5 kg of ice at

-20^{\circ}C

into water at

40^{\circ}C

? (

s_{ice} = 2100

s_{water} = 4200\ \text{J kg}^{-1}\ \text{K}^{-1}

L_f = 3.36 \times 10^{5}\ \text{J kg}^{-1}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · JEE Mains 2022 — 25 June · Q95Moderate

Example 2 · Thermal Properties of Matter · Calorimetry and Latent Heat

A copper block of mass

5.0\text{ }kg

is heated to a temperature of

500^{\circ}C

and is placed on a large ice block. What is the maximum amount of ice that can melt? [Specific heat of copper:

0.39Jg^{- 1}\ ^{\circ}C^{- 1}

and latent heat of fusion of water :

335\text{ }J{\text{ }g}^{- 1}

]

Temperature does not rise during melting

Heat supplied while ice melts goes into the change of phase. A graph that keeps rising through 0°C, or an answer that adds a temperature rise there, is wrong.

Forgetting to warm the ice to 0°C first

Ice below 0°C must reach 0°C before it can melt. That term, ms_ice times the degrees below zero, is small but changes the answer between close options.

Latent heat in kJ

Latent heats are often given in kJ/kg or cal/g while specific heats are in J kg⁻¹ K⁻¹. Convert to one unit before adding the stages.

Drill 7 more on latent heat and the heating curve

Concept 3 of 3: Mixing ice and water: heat lost equals heat gained

In an insulated mixture, every joule the hot part loses is gained by the cold part. With ice there is a catch: you do not know in advance whether all of it melts. So test first. Compare the heat the water can give out by cooling to 0°C with the heat the ice needs to reach 0°C and melt. Which one is smaller decides what the final state is.

Definition

Heat lost by hot parts $=$ heat gained by cold parts (no losses).
Test first: $Q_{water} = m_ws_w(T_w - 0)$ against $Q_{ice} = m_is_i(0 - T_i) + m_iL$ .
If $Q_{water} < Q_{ice}$ : the final temperature is $0^{\circ}C$ and only part of the ice melts; melted mass $= (Q_{water} - m_is_i|T_i|)/L$ .
If $Q_{water} > Q_{ice}$ : all the ice melts; then solve $m_ws_w(T_w - T) = Q_{ice} + m_is_wT$ for T.
Equal masses of two liquids: $s_1(T - T_1) = s_2(T_2 - T)$ , so the mass cancels and pairs of mixtures give ratios of specific heats.

Heat balance

m_ws_w(T_w - T) = m_is_i(0 - T_i) + m_iL + m_is_w(T - 0)

Worked example

50 g of ice at

0^{\circ}C

is dropped into 200 g of water at

30^{\circ}C

. Find the final temperature. (

s_w = 4.2\ \text{J g}^{-1}\ \text{K}^{-1}

L = 336\ \text{J g}^{-1}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q1Moderate

Example 3 · Thermal Properties of Matter · Calorimetry and Latent Heat

10 kg of ice at

-10^{\circ}C

is added to 100 kg of water to lower its temperature from

25^{\circ}C

. Consider no heat exchange to surroundings. The decrement to the temperature of water is

\_\_\_\_

\ ^{\circ}C

. (specific heat of ice

= 2100\text{ }J/Kg.\ ^{\circ}C

, specific heat of water

= 4200\text{ }J/Kg.\ ^{\circ}C

, latent heat of fusion of ice

= 3.36 \times10^{5}\text{ }J/Kg

)

Assuming all the ice melts

Solving the balance without the test can give a final temperature below 0°C, which is impossible with water present. Compare the two heats first; if the water's is smaller, the answer is 0°C.

Forgetting that melted ice also warms

Once the ice has melted it is water at 0°C, and it must be warmed to the final temperature too. Leave out m_i s_w T and the final temperature comes out too high.

Drill 3 more on mixing ice and water: heat lost equals heat gained

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Heat from a heater, a fuel or lost kinetic energy
Energy into heat
$Q = ms\Delta T \qquad \eta Pt = ms\Delta T \qquad f\cdot\tfrac{1}{2}mv^{2} = ms\Delta T + mL$
Latent heat and the heating curve
Latent heat and heating-curve slope
$Q = mL \qquad \frac{dT}{dQ} = \frac{1}{ms}$
Mixing ice and water: heat lost equals heat gained
Heat balance
$m_ws_w(T_w - T) = m_is_i(0 - T_i) + m_iL + m_is_w(T - 0)$

Watch out for (8)

Grams and kilograms in one equation
→ Heat from a heater, a fuel or lost kinetic energy
Stopping at the melting point
→ Heat from a heater, a fuel or lost kinetic energy
Which g the options use
→ Heat from a heater, a fuel or lost kinetic energy
Temperature does not rise during melting
→ Latent heat and the heating curve
Forgetting to warm the ice to 0°C first
→ Latent heat and the heating curve
Latent heat in kJ
→ Latent heat and the heating curve
Assuming all the ice melts
→ Mixing ice and water: heat lost equals heat gained
Forgetting that melted ice also warms
→ Mixing ice and water: heat lost equals heat gained

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