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Chemistry · Thermodynamics

Measurement of ΔU and ΔH — Calorimetry

Thermodynamics defines internal energy and enthalpy, but their numerical values reach us only through measurement. Section 5.3 of the NCERT Class 11 unit on Thermodynamics introduces calorimetry — the experimental technique that converts a measured temperature change into a heat quantity, and from there into $\Delta U$ or $\Delta H$. For NEET, this subtopic supplies the apparatus logic behind $q_V = \Delta U$ and $q_p = \Delta H$, the working equation $q = C\Delta T$, and the $\Delta H = \Delta U + \Delta n_g RT$ bridge that recurs every year.

What Calorimetry Measures

Energy changes accompanying chemical or physical processes are measured experimentally by calorimetry. The process is carried out in a vessel called a calorimeter, which is immersed in a known volume of a liquid — usually water. If the heat capacity of that liquid and the heat capacity of the calorimeter are known, the heat evolved or absorbed can be determined simply by tracking the temperature change of the assembly.

The NCERT text fixes one decisive idea: measurements are made under two different conditions, and the condition selects which thermodynamic quantity you obtain. At constant volume the heat exchanged is $q_V$, and at constant pressure it is $q_p$. These are not arbitrary labels — they connect directly to the first law of thermodynamics.

ConditionHeat symbolWhat it equalsWhy
Constant volume ($\Delta V = 0$)$q_V$$\Delta U$No expansion work, so $w = 0$
Constant pressure$q_p$$\Delta H$$H \equiv U + pV$ absorbs the $p\Delta V$ work

The first law states $\Delta U = q + w$, where $w$ is the work done on the system. Holding the volume constant removes expansion work entirely, leaving the clean identity $\Delta U = q_V$. Holding the pressure constant instead routes the heat into the enthalpy, $q_p = \Delta H$. Calorimetry is, in essence, the art of building an apparatus that enforces one of these two conditions so that the measured heat is exactly the quantity wanted.

Heat Capacity and q = CΔT

The quantitative link between a thermometer reading and a heat quantity is the heat capacity. A substance with a large heat capacity requires a lot of energy to raise its temperature. For a calorimeter assembly the relevant quantity is the heat capacity $C$ of the whole device — vessel, water bath, stirrer and thermometer combined — measured in $\text{J K}^{-1}$ or $\text{kJ K}^{-1}$.

Once $C$ is known, the heat exchanged for any measured temperature change $\Delta T$ follows from the central calorimetric equation:

$$ q = C\,\Delta T $$

Figure 1 System reaction mixture loses q Calorimeter heat capacity C temperature rises by ΔT heat q gained = C ΔT → lost by system = − C ΔT

Heat released by the reacting system is captured by the calorimeter. Because heat lost by the system equals heat gained by the calorimeter, the two have equal magnitude but opposite signs: $q_{\text{system}} = -C\Delta T$.

The sign convention is worth pinning down. The calorimeter gains heat $+C\Delta T$, so the reacting system must have lost the same amount, $q_{\text{system}} = -C\Delta T$. For an exothermic reaction the temperature rises ($\Delta T > 0$), and $q_{\text{system}}$ comes out negative — exactly as it should, since heat has flowed out of the system.

ΔU at Constant Volume — Bomb Calorimeter

For chemical reactions, the heat absorbed at constant volume is measured in a bomb calorimeter. A heavy steel vessel — the bomb — is immersed in a stirred water bath, and the whole device is the calorimeter. The steel vessel sits in the water bath specifically to ensure that no heat is lost to the surroundings.

A combustible substance is burnt in pure dioxygen supplied inside the steel bomb. The heat released during the reaction is transferred to the water around the bomb, and the water's temperature is monitored. Because the bomb is sealed, its volume does not change, so the energy change is measured at constant volume. Crucially, no work is done even when gases are involved, because $\Delta V = 0$ inside the rigid vessel. Under these conditions the measured heat is $q_V$, which converts to $\Delta U$ through the known heat capacity using $q = C\Delta T$.

Figure 2 Insulated jacket (no heat loss) water bath steel bomb sample + O₂ thermometer stirrer Volume fixed → w = 0 → q_V = ΔU

Bomb calorimeter (NCERT Fig. 5.7). The sealed steel bomb keeps the volume constant, so no expansion work is done and the heat transferred to the surrounding water gives $q_V = \Delta U$.

NEET Trap

A bomb calorimeter measures ΔU, not ΔH

Examiners frequently describe combustion "in a bomb calorimeter" and then ask for the enthalpy of combustion. The bomb works at constant volume, so it gives $q_V = \Delta U$ directly. The enthalpy must be obtained afterwards through $\Delta H = \Delta U + \Delta n_g RT$.

Bomb (sealed, rigid) → $q_V = \Delta U$. Open vessel (atmospheric) → $q_p = \Delta H$.

ΔH at Constant Pressure — Coffee-Cup Calorimeter

Most chemical reactions of interest are not carried out at constant volume but in flasks or test tubes open to the atmosphere, that is, at constant pressure. To measure these, a constant-pressure calorimeter is used — at the school-laboratory level a simple insulated "coffee-cup" vessel open to the atmosphere. Because the pressure stays constant (atmospheric), the heat absorbed or evolved is $q_p$, and we know that $\Delta H = q_p$ at constant pressure.

This heat at constant pressure is also called the heat of reaction or enthalpy of reaction, $\Delta_r H$. In an exothermic reaction the system loses heat to the surroundings, so $q_p$ is negative and $\Delta_r H$ is negative; in an endothermic reaction heat is absorbed, $q_p$ is positive and $\Delta_r H$ is positive.

Figure 3 open to atmosphere → p constant insulating cup reaction solution thermometer stirrer Pressure fixed → q_p = ΔH = Δ_r H

Constant-pressure (coffee-cup) calorimeter, NCERT Fig. 5.8. Open to the atmosphere, it measures $q_p$, which equals the reaction enthalpy $\Delta_r H$.

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Build the foundation

The whole apparatus logic rests on the definitions of $C$, $C_p$ and $C_V$. Revise Enthalpy and Heat Capacity before you tackle calorimetric problems.

Bridging ΔU and ΔH

Since enthalpy is defined as $H = U + pV$, a change at constant pressure gives $\Delta H = \Delta U + p\Delta V$. For a reaction involving ideal gases, the ideal gas law lets us replace $p\Delta V$ by a tidy expression. Writing $pV_A = n_A RT$ for the gaseous reactants and $pV_B = n_B RT$ for the gaseous products at constant temperature and pressure,

$$ p\,\Delta V = p(V_B - V_A) = (n_B - n_A)RT = \Delta n_g RT $$

where $\Delta n_g$ is the number of moles of gaseous products minus the number of moles of gaseous reactants. Substituting gives the bridge equation:

$$ \Delta H = \Delta U + \Delta n_g RT $$

This is the equation that turns a bomb-calorimeter $\Delta U$ into the enthalpy $\Delta H$, and vice versa. When no gaseous moles change ($\Delta n_g = 0$) the two are identical, $\Delta H = \Delta U$. The difference is also negligible for systems of only solids and liquids, because such phases barely change volume on heating; it becomes significant only when the number of gas-phase moles changes.

Reaction$\Delta n_g$Relation
$\ce{C(graphite) + O2(g) -> CO2(g)}$$1 - 1 = 0$$\Delta H = \Delta U$
$\ce{H2(g) + 1/2 O2(g) -> H2O(l)}$$0 - 1.5 = -1.5$$\Delta H = \Delta U - 1.5RT$
$\ce{CaCO3(s) -> CaO(s) + CO2(g)}$$1 - 0 = +1$$\Delta H = \Delta U + RT$

Worked Examples

Worked Example 1 · NCERT-type

$1\,\text{g}$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298\ \text{K}$ and $1$ atm:

$$\ce{C(graphite) + O2(g) -> CO2(g)}$$

During the reaction the temperature rises from $298\ \text{K}$ to $299\ \text{K}$. If the heat capacity of the bomb calorimeter is $20.7\ \text{kJ K}^{-1}$, what is the enthalpy change for the reaction at $298\ \text{K}$ and $1$ atm?

Step 1 — heat gained by the calorimeter. With $C_V$ the heat capacity, the calorimeter absorbs $q = C_V\,\Delta T$.

Step 2 — heat from the reaction. Heat lost by the system equals heat gained by the calorimeter, so it has the same magnitude but the opposite sign:

$$ q = -\,C_V\,\Delta T = -\,20.7\ \text{kJ K}^{-1} \times (299 - 298)\ \text{K} = -20.7\ \text{kJ} $$

The negative sign confirms the reaction is exothermic. Thus $\Delta U$ for combustion of $1\,\text{g}$ of graphite is $-20.7\ \text{kJ}$.

Step 3 — per mole. The molar mass of carbon is $12.0\ \text{g mol}^{-1}$, so for $1$ mole:

$$ \Delta U = \frac{-20.7\ \text{kJ}}{1\ \text{g}} \times 12.0\ \text{g mol}^{-1} = -2.48 \times 10^{2}\ \text{kJ mol}^{-1} $$

Step 4 — convert to ΔH. Here $\Delta n_g = 1 - 1 = 0$, so $\Delta H = \Delta U = -2.48 \times 10^{2}\ \text{kJ mol}^{-1}$.

Worked Example 2 · NCERT-type

If water vapour is assumed to be a perfect gas, the molar enthalpy change for vaporisation of $1$ mol of water at $1$ bar and $100\,^{\circ}\text{C}$ is $41\ \text{kJ mol}^{-1}$. Calculate the internal energy change when (i) $1$ mol of water is vaporised at $1$ bar pressure and $100\,^{\circ}\text{C}$.

$$\ce{H2O(l) -> H2O(g)}$$

Step 1 — rearrange the bridge equation. From $\Delta H = \Delta U + \Delta n_g RT$,

$$ \Delta U = \Delta H - \Delta n_g RT $$

Step 2 — find $\Delta n_g$. One mole of gaseous water is produced from liquid water, so $\Delta n_g = +1$.

Step 3 — substitute. With $T = 373\ \text{K}$ and $R = 8.314\times10^{-3}\ \text{kJ K}^{-1}\text{mol}^{-1}$:

$$ \Delta U = 41\ \text{kJ mol}^{-1} - (1)(8.314\times10^{-3})(373)\ \text{kJ mol}^{-1} $$

$$ \Delta U = 41 - 3.10 = 37.9\ \text{kJ mol}^{-1} $$

The internal energy change is smaller than the enthalpy change because part of the $41\ \text{kJ}$ goes into the expansion work done by the newly formed vapour against the atmosphere.

NEET Trap

Keep the units of R consistent

In $\Delta H = \Delta U + \Delta n_g RT$, if energies are in kJ then $R$ must be $8.314\times10^{-3}\ \text{kJ K}^{-1}\text{mol}^{-1}$. Mixing $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ with kJ energies inflates the correction term a thousandfold — a classic silent error.

Energies in kJ → $R = 8.314\times10^{-3}\ \text{kJ K}^{-1}\text{mol}^{-1}$. Count only gaseous species in $\Delta n_g$.

Constant-V vs Constant-p — Summary

The two measurement modes are best contrasted side by side. The choice of apparatus is dictated entirely by whether the volume or the pressure is held fixed, and this in turn fixes whether the measured heat is $\Delta U$ or $\Delta H$.

FeatureConstant volumeConstant pressure
ApparatusBomb calorimeter (sealed steel bomb)Coffee-cup / open calorimeter
Held fixedVolume ($\Delta V = 0$)Pressure (atmospheric)
Expansion work$w = 0$$w = -p\Delta V$
Heat measured$q_V$$q_p$
Quantity obtained$\Delta U = q_V$$\Delta H = q_p = \Delta_r H$
Working relation$q_V = C_V\,\Delta T$$q_p = C_p\,\Delta T$
Conversion$\Delta H = \Delta U + \Delta n_g RT$

Whichever route is used, the experimental skeleton is identical: calibrate the calorimeter to know its heat capacity, run the process, read the temperature change, and apply $q = C\Delta T$. The thermodynamics enters only in deciding what that $q$ represents and, if necessary, converting between $\Delta U$ and $\Delta H$.

Quick Recap

Calorimetry in one screen

  • Calorimetry measures heat from a temperature change using $q = C\Delta T$, where $C$ is the heat capacity of the calorimeter assembly.
  • At constant volume $w = 0$, so $q_V = \Delta U$ — measured in a sealed bomb calorimeter.
  • At constant pressure $q_p = \Delta H = \Delta_r H$ — measured in an open constant-pressure (coffee-cup) calorimeter.
  • Heat lost by the system equals heat gained by the calorimeter: $q_{\text{system}} = -C\Delta T$ (negative ⇒ exothermic).
  • Convert between the two with $\Delta H = \Delta U + \Delta n_g RT$; when $\Delta n_g = 0$, $\Delta H = \Delta U$.
  • $\Delta H \approx \Delta U$ for solids and liquids; the difference matters only when gaseous moles change.

NEET PYQ Snapshot — Measurement of ΔU and ΔH — Calorimetry

Real NEET items on the ΔU↔ΔH bridge and the heat-capacity relations that calorimetry rests on.

NEET 2023

Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?

  • (1) $\Delta H + \Delta U = \Delta n_g R$
  • (2) $\Delta H = \Delta U - \Delta n_g RT$
  • (3) $\Delta H = \Delta U + \Delta n_g RT$
  • (4) $\Delta H - \Delta U = -\Delta n_g RT$
Answer: (3)

From $H = U + pV$, at constant pressure $\Delta H = \Delta U + p\Delta V$. For ideal gases $p\Delta V = \Delta n_g RT$, giving $\Delta H = \Delta U + \Delta n_g RT$ — the exact relation used to convert a bomb-calorimeter $\Delta U$ into $\Delta H$.

NEET 2021

Which one among the following is the correct option for right relationship between $C_P$ and $C_V$ for one mole of ideal gas?

  • (1) $C_V = R\,C_P$
  • (2) $C_P + C_V = R$
  • (3) $C_P - C_V = R$
  • (4) $C_P = R\,C_V$
Answer: (3)

At constant volume $q_V = C_V\Delta T = \Delta U$; at constant pressure $q_p = C_P\Delta T = \Delta H$. Since $\Delta H = \Delta U + R\Delta T$ for one mole of ideal gas, $C_P\Delta T = C_V\Delta T + R\Delta T$, hence $C_P - C_V = R$. This links the two heat capacities a calorimeter uses.

NEET 2017

A gas is allowed to expand in a well insulated container against a constant external pressure of $2.5$ atm from an initial volume of $2.50\,\text{L}$ to a final volume of $4.50\,\text{L}$. The change in internal energy $\Delta U$ of the gas in joules will be:

  • (1) $+505\,\text{J}$
  • (2) $1136.25\,\text{J}$
  • (3) $-500\,\text{J}$
  • (4) $-505\,\text{J}$
Answer: (4)

Insulated container ⇒ $q = 0$. So $\Delta U = w = -p_{ext}\Delta V = -2.5(4.50-2.50) = -5\ \text{L atm}$. Using $1\ \text{L atm} \approx 101\ \text{J}$: $\Delta U = -5 \times 101 = -505\ \text{J}$. The link to calorimetry: only at constant volume would $\Delta U$ equal $q$; here it equals the work.

Concept

In a bomb calorimeter of heat capacity $5.0\ \text{kJ K}^{-1}$, a sample is burnt at constant volume and the temperature rises by $2.0\ \text{K}$. What is $\Delta U$ for the process?

  • (1) $+10.0\,\text{kJ}$
  • (2) $-10.0\,\text{kJ}$
  • (3) $+2.5\,\text{kJ}$
  • (4) $-2.5\,\text{kJ}$
Answer: (2)

Heat gained by the calorimeter is $C\Delta T = 5.0 \times 2.0 = 10.0\ \text{kJ}$. The system lost the same amount, so $q_{\text{system}} = -10.0\ \text{kJ}$. At constant volume $\Delta U = q_V = -10.0\ \text{kJ}$ — exothermic, as the negative sign and temperature rise confirm.

FAQs — Measurement of ΔU and ΔH — Calorimetry

The recurring conceptual snags students hit when reading temperature changes as heat.

Why is ΔU measured at constant volume and ΔH at constant pressure?

From the first law, ΔU = q + w. At constant volume there is no expansion work (ΔV = 0, so w = 0), which leaves ΔU = qV — the heat exchanged at constant volume equals the internal energy change exactly. This is what a sealed bomb calorimeter measures. At constant pressure the system can do expansion work, and the heat absorbed becomes the enthalpy change, qp = ΔH, because H is defined as H = U + pV. So the apparatus is chosen to match the quantity you want: a rigid bomb for qV = ΔU, an open constant-pressure vessel for qp = ΔH.

What is the heat capacity of a calorimeter and how is it used?

The heat capacity of a calorimeter, C, is the heat needed to raise the temperature of the whole assembly (vessel, water bath, thermometer, stirrer) by one kelvin, with units of J K⁻¹ or kJ K⁻¹. Once C is known, any measured temperature rise ΔT converts directly into heat through q = CΔT. The calorimeter is usually calibrated beforehand by supplying a known amount of electrical heat or by burning a substance of known enthalpy of combustion.

How do you convert ΔU measured in a bomb calorimeter into ΔH?

Use ΔH = ΔU + ΔngRT, where Δng is the number of moles of gaseous products minus the number of moles of gaseous reactants, R is the gas constant and T is the temperature in kelvin. This follows because at constant pressure pΔV = ΔngRT for ideal gases. When no gases change in number (Δng = 0), as in the combustion of graphite to carbon dioxide, ΔH = ΔU.

Why does heat from the reaction get a negative sign in q = -CΔT?

Heat lost by the reaction mixture is gained by the calorimeter, so the two have the same magnitude but opposite signs. The calorimeter absorbs +CΔT, hence the reaction releases -CΔT. For an exothermic reaction the temperature rises, ΔT is positive, and q for the system comes out negative — correctly signalling that heat flowed out of the system.

Why is the bomb calorimeter immersed in a water bath?

The steel bomb is immersed in a stirred water bath so that virtually no heat escapes to the surroundings. Heat released by the burning sample is transferred to the surrounding water, and its temperature rise is monitored. Because the assembly is well insulated, almost all the reaction heat is accounted for by the temperature change of the calorimeter, allowing an accurate qV measurement.

Is the difference between ΔH and ΔU important for reactions of only solids and liquids?

No. Solids and liquids undergo negligible volume change on heating, so pΔV is tiny and ΔH ≈ ΔU for such systems. The difference becomes significant only when gases are produced or consumed, because the gas-phase volume change makes the pΔV = ΔngRT term appreciable.

Related Topics
Thermodynamics Back to the full chapter hub. First Law of Thermodynamics ΔU = q + w — the basis for qV = ΔU. Enthalpy and Heat Capacity H = U + pV, plus Cp, Cv and Cp − Cv = R. Enthalpies of Reactions ΔrH and the standard enthalpy types calorimetry yields. Hess's Law Combine measured enthalpies along any path.