What is the principle of calorimetry?

In an isolated system, where no heat escapes to the surroundings, the heat lost by the bodies at higher temperature equals the heat gained by the bodies at lower temperature. This single energy-conservation statement, heat lost = heat gained, is the principle of calorimetry and is the only equation you need to set up any mixing problem.

What is the water equivalent of a calorimeter?

The water equivalent w of a calorimeter is the mass of water that would absorb the same amount of heat as the calorimeter for the same temperature rise. If the calorimeter has mass m and specific heat s, then w = m·s/s_water. Adding this 'extra mass of water' to the real water lets you treat the calorimeter and its contents as one body.

Why must the calorimeter itself be included in the heat balance?

The metallic vessel and stirrer are in thermal contact with the water and rise (or fall) in temperature along with it, so they absorb (or release) heat too. Ignoring the calorimeter term m·s·ΔT systematically biases the measured specific heat. NEET frequently rewards candidates who remember to add the calorimeter or its water equivalent.

How do I handle a phase change inside a mixing problem?

Add a latent-heat term m·L at the temperature where the change of state occurs (0 °C for ice–water, 100 °C for water–steam), because the substance absorbs or releases heat at constant temperature during the change. First test whether the available heat is enough to complete the phase change; if not, the final temperature is pinned at the transition temperature and a mixture of two phases remains.

Can the final temperature lie outside the range of the two starting temperatures?

No. With no external heat source, the equilibrium temperature must lie between the lowest and highest starting temperatures. A final temperature above the hot body or below the cold body violates energy conservation. If algebra hands you such a value, a phase change has been overlooked and the true answer is pinned at a transition temperature.

Why is water used as a coolant and to measure heat?

Water has the highest specific heat capacity of common substances, 4186 J kg⁻¹ K⁻¹, so it absorbs a large amount of heat for a small temperature rise. This makes it ideal both as a coolant in radiators and as the working fluid in the method of mixtures, where its well-known specific heat sets the scale against which an unknown is measured.

Calorimetry — NEET Physics

The principle of calorimetry

A system is isolated when no heat is exchanged with its surroundings. NCERT states the governing rule directly: when different parts of an isolated system sit at different temperatures, heat transfers from the hotter part to the colder part, and the heat lost by the part at higher temperature is exactly equal to the heat gained by the part at lower temperature. Bring a hot body into contact with a cold one inside such a system and they drift toward a common equilibrium temperature, exchanging heat but losing none to the outside.

This is nothing more than conservation of energy applied to heat. It is the only equation calorimetry needs. Every mixing problem, every specific-heat measurement, every ice-into-water puzzle reduces to balancing one column of heat-lost terms against one column of heat-gained terms.

Heat LOST by hot bodies	=	Heat GAINED by cold bodies
Each hot body cooling: $m\,s\,(T_{\text{hot}}-T_f)$	=	Each cold body warming: $m\,s\,(T_f-T_{\text{cold}})$
Liquid freezing / vapour condensing: $m\,L$	=	Solid melting / liquid vaporising: $m\,L$
Always positive — releasing heat	=	Always positive — absorbing heat

The quantity $Q = m\,s\,\Delta T$ is the sensible heat that changes temperature without changing state; the term $Q = m\,L$ is the latent heat absorbed or released at constant temperature during a change of state. Keeping these two kinds of term in separate but parallel columns is the discipline that makes calorimetry mechanical.

The calorimeter: a copper vessel and matching stirrer sit inside a wooden jacket packed with glass wool, with a thermometer through the lid. The jacket reduces heat exchange with the surroundings so the contents behave as an isolated system.

The calorimeter and water equivalent

A calorimeter is the device in which heat measurement is done. NCERT describes it as a metallic vessel — copper or aluminium — with a matching stirrer, kept inside a wooden jacket filled with insulating material such as glass wool. The jacket is the heat shield: it reduces heat loss from the inner vessel so the contents approximate an isolated system, and a mercury thermometer reaches the interior through an opening in the lid.

The catch that NEET exploits is that the vessel and stirrer are themselves in thermal contact with the liquid. They warm up and cool down with it, so they absorb and release heat too. A correct heat balance must include a term $m_c s_c \Delta T$ for the calorimeter alongside the term for the water it holds.

It is often tidier to fold the calorimeter into the water using its water equivalent $w$ — the mass of water that would absorb the same heat as the calorimeter for the same temperature change.

The water equivalent of a body of mass $m_c$ and specific heat $s_c$ is

$$ w = \frac{m_c\,s_c}{s_{\text{water}}} \qquad\text{so that}\qquad m_c\,s_c\,\Delta T = w\,s_{\text{water}}\,\Delta T. $$

Replacing the calorimeter with its water equivalent lets you treat "water + calorimeter" as a single effective mass $(m_w + w)$ of water. The two descriptions are identical; choose whichever keeps the algebra cleaner.

Setting up the mixing equation

When two bodies at temperatures $T_1 > T_2$ are mixed and reach equilibrium at $T_f$, with no phase change, the principle of calorimetry gives a single equation that can be solved for $T_f$. Follow the same fixed sequence every time.

Step	Action	Why it matters
1	List every body and its mass, specific heat and starting temperature, including the calorimeter.	A missing body (usually the calorimeter) is the commonest source of error.
2	Decide which bodies cool (lose heat) and which warm (gain heat).	Sets the sign of each $\Delta T$ and avoids negative answers.
3	Write heat lost $= \sum m\,s\,(T_{\text{start}}-T_f)$ for hot bodies.	Each term is positive when written as (start − final).
4	Write heat gained $= \sum m\,s\,(T_f-T_{\text{start}})$ for cold bodies.	Each term is positive when written as (final − start).
5	Equate the two columns and solve for the single unknown $T_f$.	One equation, one unknown — provided no phase change intervenes.

For a hot body of mass $m_1$, specific heat $s_1$ mixed with a cold body $(m_2, s_2)$, with no phase change, the balance is

$$ m_1 s_1 (T_1 - T_f) = m_2 s_2 (T_f - T_2), $$

which rearranges to the weighted-average form

$$ T_f = \frac{m_1 s_1 T_1 + m_2 s_2 T_2}{m_1 s_1 + m_2 s_2}. $$

The product $m s$ is the body's heat capacity; the final temperature is simply the heat-capacity-weighted average of the starting temperatures. Because it is a weighted average, $T_f$ must lie between $T_1$ and $T_2$ — a fact that doubles as a sanity check.

Heat-balance bar. The heat released as the hot body cools to $T_f$ (red) equals the heat absorbed as the cold liquid and the calorimeter warm to $T_f$ (teal). The equilibrium temperature always lies between the two starting temperatures.

Method of mixtures — measuring specific heat

The most important laboratory application is the method of mixtures: drop a heated solid of unknown specific heat into water in a calorimeter and measure the final temperature. Because water's specific heat is known accurately (4186 J kg⁻¹ K⁻¹, the highest of common substances), it sets the reference scale against which the unknown is read off.

Foundation

The whole method rests on $Q = m\,s\,\Delta T$. If that relation is shaky, revisit specific heat capacity before working the balance equations below.

NCERT Example 10.3 — specific heat of aluminium

NCERT Example 10.3

A sphere of 0.047 kg aluminium is kept in boiling water until it reaches 100 °C, then transferred to a 0.14 kg copper calorimeter holding 0.25 kg water at 20 °C. The mixture settles at 23 °C. Find the specific heat of aluminium. ($s_w = 4.18\times10^{3}$, $s_{\text{Cu}} = 0.386\times10^{3}$ J kg⁻¹ K⁻¹.)

Identify hot and cold. The aluminium sphere is the only hot body: it cools from 100 °C to 23 °C, a drop of $\Delta T_1 = 77$ °C. The water and the copper calorimeter are cold: both warm from 20 °C to 23 °C, a rise of $\Delta T_2 = 3$ °C.

Heat lost by aluminium. $Q_{\text{lost}} = m_{\text{Al}}\,s_{\text{Al}}\,\Delta T_1 = 0.047 \times s_{\text{Al}} \times 77$.

Heat gained by water + calorimeter. $Q_{\text{gained}} = (m_w s_w + m_c s_{\text{Cu}})\,\Delta T_2 = (0.25\times4.18\times10^{3} + 0.14\times0.386\times10^{3})\times 3.$

Equate and solve. Setting $Q_{\text{lost}} = Q_{\text{gained}}$: $$ 0.047\,s_{\text{Al}}\,(77) = \big(1045 + 54.0\big)(3), $$ giving $s_{\text{Al}} \approx 0.911\times10^{3}$ J kg⁻¹ K⁻¹, i.e. $0.911$ kJ kg⁻¹ K⁻¹ — close to the tabulated 900 J kg⁻¹ K⁻¹.

Why the calorimeter matters. Drop the copper term and the right side falls from $\approx 3297$ J to $3135$ J, inflating $s_{\text{Al}}$ by about 5 %. The small $0.14\times0.386\times10^{3}$ contribution is exactly the kind of term NEET checks for.

Worked mixing problem — two liquids

Worked example

200 g of water at 80 °C is poured into 300 g of water at 20 °C in a calorimeter of water equivalent 50 g. Find the equilibrium temperature. (For water, $s$ is the same throughout and cancels.)

Identify hot and cold. Hot: 200 g of water cooling from 80 °C. Cold: 300 g of water warming from 20 °C, together with the calorimeter (water equivalent 50 g, also starting at 20 °C, so effectively $300+50 = 350$ g of cold water).

Balance (the common $s_w$ cancels). $$ 200\,(80 - T_f) = 350\,(T_f - 20). $$

Solve. $16000 - 200 T_f = 350 T_f - 7000 \Rightarrow 550\,T_f = 23000 \Rightarrow T_f \approx 41.8$ °C.

Sanity check. 41.8 °C lies between 20 °C and 80 °C, as a weighted average must. Had the calorimeter been ignored, the answer would have risen to 44 °C — the calorimeter pulls the equilibrium toward the cold side.

Calorimetry with a phase change

The hardest NEET calorimetry questions combine mixing with a change of state. When a substance melts, freezes, vaporises or condenses, it absorbs or releases latent heat $m L$ at constant temperature — the thermometer holds steady at 0 °C or 100 °C while the heat goes into rearranging the molecules rather than speeding them up. The latent-heat term joins the heat balance as an extra entry in the appropriate column.

The non-obvious part is that you cannot assume the phase change goes to completion. The right procedure is to test the heat budget at the transition temperature before solving for any final temperature.

Situation at the transition temperature	What happens	Final state
Heat available exceeds the latent heat needed	Phase change completes; leftover heat changes the temperature further	Single phase, $T_f$ found from a fresh balance
Heat available exactly equals the latent heat	Phase change just completes	Single phase, exactly at the transition temperature
Heat available is less than the latent heat needed	Phase change is only partial; temperature cannot move	Two phases coexist, $T_f$ pinned at the transition temperature

The latent heats and specific heats needed for these problems link directly to the change-of-state framework, so keep both sets of constants at hand.

Related drill

The $m L$ terms come from latent heat — see change of state & latent heat for fusion and vaporisation values and the flat segments of the heating curve.

Worked phase-change example

NCERT Example 10.5

Calculate the heat required to convert 3 kg of ice at −12 °C, kept in a calorimeter, to steam at 100 °C at atmospheric pressure. ($s_{\text{ice}} = 2100$, $s_{\text{water}} = 4186$ J kg⁻¹ K⁻¹; $L_f = 3.35\times10^{5}$, $L_v = 2.256\times10^{6}$ J kg⁻¹.)

Break the path into four legs, two sensible and two latent. The temperature climbs, pauses at 0 °C while ice melts, climbs again, then pauses at 100 °C while water boils.

$Q_1$ — warm ice −12 °C → 0 °C: $m s_{\text{ice}} \Delta T = 3\times2100\times12 = 75\,600$ J.

$Q_2$ — melt ice at 0 °C: $m L_f = 3\times3.35\times10^{5} = 1\,005\,000$ J.

$Q_3$ — warm water 0 °C → 100 °C: $m s_{\text{water}} \Delta T = 3\times4186\times100 = 1\,255\,800$ J.

$Q_4$ — boil water at 100 °C: $m L_v = 3\times2.256\times10^{6} = 6\,768\,000$ J.

Total: $Q = Q_1+Q_2+Q_3+Q_4 \approx 9.1\times10^{6}$ J. The two latent terms dominate — the bulk of the energy goes into boiling, not heating.

This worked path is the template for the mixing version. If instead a fixed amount of hot water (or steam) supplied this heat, you would equate the supplier's heat-lost to whichever of $Q_1,\dots,Q_4$ the budget could actually fund, stopping at the leg where the heat runs out and pinning $T_f$ there.

Quick recap

Calorimetry in one breath

Principle: in an isolated system, heat lost by hot bodies = heat gained by cold bodies.
Sensible heat $Q = m s \Delta T$; latent heat $Q = m L$ at constant temperature during a change of state.
Always include the calorimeter — directly as $m_c s_c \Delta T$ or via its water equivalent $w = m_c s_c / s_w$.
No-phase-change mixing: $T_f = \dfrac{m_1 s_1 T_1 + m_2 s_2 T_2}{m_1 s_1 + m_2 s_2}$, the heat-capacity-weighted average, always between the two starting temperatures.
With a phase change, test the heat budget at 0 °C or 100 °C first; if the latent heat cannot be met, $T_f$ is pinned at the transition and two phases coexist.
Method of mixtures measures an unknown specific heat against water, whose $s = 4186$ J kg⁻¹ K⁻¹ is the highest of common substances.

Calorimetry