Why does a bomb calorimeter measure Delta U and not Delta H?

The bomb has a fixed steel volume, so combustion occurs at constant V. With no volume change the work is zero, and the first law gives heat equal to Delta U. To get Delta H use Delta H = Delta U + Delta n_g RT.

Why does a coffee-cup calorimeter measure Delta H directly?

The cup is open to atmosphere so the reaction runs at constant pressure. Under constant P, the heat exchanged equals Delta H directly, with no correction needed. It is the standard tool for solution-phase enthalpies of neutralization, dissolution and dilution.

Why is Cv for a monoatomic ideal gas exactly three-halves R?

A monoatomic gas has only 3 translational degrees of freedom. Each contributes half kT per molecule by equipartition, so U = three-halves nRT and Cv = three-halves R. Mayer's relation gives Cp = five-halves R.

In a bomb calorimeter, why is oxygen loaded at high pressure?

To ensure complete combustion of the sample. Limited oxygen could give partial combustion (CO instead of CO2), producing an incorrect heat of combustion. Loading O2 at 20 to 30 bar guarantees a large excess.

Calorimetry

Q: Why is Cp greater than Cv for a gas?

At constant pressure some heat goes into expansion work, so more heat is needed to achieve the same temperature rise as at constant volume. The difference Cp - Cv equals R for an ideal gas, exactly the work done per unit temperature per mole.

Q: What is the water equivalent of a calorimeter?

It is the mass of water that would absorb the same heat as the calorimeter for the same temperature rise. Adding it to the actual water mass lets you treat the whole system as one equivalent mass of water, simplifying q = m s Delta T.

Q: Why is water used as the working fluid in calorimeters?

Water has a very high specific heat capacity (about 4.184 J per g per K). A small amount of water absorbs a lot of heat for a small temperature rise, reducing errors from heat loss to the surroundings and improving accuracy.

Q: Can heat capacity be negative or infinite? Give examples.

Yes - heat capacity is path-dependent. Isothermal processes have Delta T = 0 with nonzero q, giving infinite C. Adiabatic processes have q = 0 with nonzero Delta T, giving zero C. Certain unusual processes can even give negative C.

ChemistryChemical ThermodynamicsFor JEE aspirants

Calorimetry is the experimental measurement of heat exchanged during physical and chemical changes. The core idea is simple: measure the temperature change $Δ T$ of a known mass, multiply by an appropriate heat capacity, and you get the heat exchanged. This concept covers the three types of heat capacity (total, molar, specific), the crucial relation $C_{p} - C_{v} = R$ (Mayer's relation) for an ideal gas, and the two most important instruments in JEE and NEET chemistry - the bomb calorimeter (constant volume, measures $Δ U$ ) and the coffee-cup calorimeter (constant pressure, measures $Δ H$ ).

Key Formulas - Quick Reference

Total heat capacity: $q = C_{T} Δ T$ (units: J K $^{- 1}$ )
Molar heat capacity: $q = n C Δ T$ (units: J mol $^{- 1}$ K $^{- 1}$ )
Specific heat capacity: $q = m s Δ T$ (units: J g $^{- 1}$ K $^{- 1}$ )
Relations: $C_{T} = n C = m s$ ; $C = M s$ where $M$ is molar mass
Mayer's relation (ideal gas): $C_{p} - C_{v} = R$
Ratio of heat capacities: $γ = C_{p} / C_{v}$
Ideal gas heat capacities: $C_{v} = \frac{f}{2} R$ , $C_{p} = (\frac{f}{2} + 1) R$
Bomb calorimeter: $q_{v} = Δ U = C_{cal} Δ T$ ; then $Δ H = Δ U + Δ n_{g} R T$
Coffee-cup calorimeter: $q_{p} = Δ H = - (m_{water} s_{water} + C_{cal}) Δ T$

1. What is heat capacity

The heat capacity of a sample is the quantity of heat required to raise its temperature by one degree (Celsius or Kelvin - both are the same for a temperature change).

q = C Δ T

or in differential form

C = \frac{d q}{d T}

Because heat itself is a path function, heat capacity depends on the process (constant $P$ , constant $V$ , etc.) - not just on the substance. In JEE/NEET the two important process-specific heat capacities are $C_{p}$ (constant pressure) and $C_{v}$ (constant volume).

2. Three ways to express heat capacity

Total heat capacity $(C_{T})$

Heat required to raise the temperature of the whole sample by $1 K$ :

q = C_{T} Δ T, C_{T}

in J K

^{- 1}

Extensive property (doubles if the sample doubles). Path-function.

Molar heat capacity $(C)$

Heat required to raise the temperature of one mole of substance by $1 K$ :

q = n C Δ T, C

in J mol

^{- 1}

^{- 1}

Intensive property. For an ideal gas, $C_{p}$ and $C_{v}$ are intensive.

Specific heat capacity $(s)$

Heat required to raise the temperature of one gram of substance by $1 K$ :

q = m s Δ T, s

in J g

^{- 1}

^{- 1}

Intensive property. Water has $s = 4.184 J g^{- 1} K^{- 1}$ - very high, which is why water is used in calorimeters.

Connecting the three

C_{T} = n C = m s

and

C = M s

where $M$ is the molar mass and $n = m / M$ .

3. $C_{p}$ and $C_{v}$ - heat capacity depends on the process

When the same mole of gas is heated at constant volume vs constant pressure, more heat is needed at constant pressure - because the gas also has to do expansion work. Two important cases:

Constant volume: heat = $Δ U$

If volume is constant, $d V = 0$ so $w = - P d V = 0$ . By the first law $d U = d q_{v}$ , hence:

d U = n C_{v} d T, Δ U = n C_{v} (T_{2} - T_{1})

Thus $C_{v} = (\frac{\partial U}{\partial T})_{V} \cdot \frac{1}{n}$ .

Constant pressure: heat = $Δ H$

If pressure is constant, we have to use enthalpy $H = U + P V$ . Then $d H = d U + P d V = d q_{p}$ :

d H = n C_{p} d T, Δ H = n C_{p} (T_{2} - T_{1})

Thus $C_{p} = (\frac{\partial H}{\partial T})_{P} \cdot \frac{1}{n}$ .

4. Mayer's relation: $C_{p} - C_{v} = R$

For an ideal gas, $H = U + P V = U + n R T$ . Differentiating with respect to $T$ at constant pressure:

$\frac{d H}{d T} = \frac{d U}{d T} + n R$

Dividing by $n$ :

C_{p} - C_{v} = R

(Mayer's relation, ideal gas only)

The ratio $γ = C_{p} / C_{v}$ is called the adiabatic index. Common values:

Type of gas	Degrees of freedom $f$	$C_{v}$	$C_{p}$	$γ = C_{p} / C_{v}$
Monoatomic (He, Ar)	3	$\frac{3}{2} R$	$\frac{5}{2} R$	$5/3 \approx 1.67$
Diatomic (N $_{2}$ , O $_{2}$ , H $_{2}$ )	5	$\frac{5}{2} R$	$\frac{7}{2} R$	$7/5 = 1.40$
Non-linear polyatomic	6	$3 R$	$4 R$	$4/3 \approx 1.33$

Origin of $C_{v} = \frac{f}{2} R$ . By the equipartition of energy, each active degree of freedom contributes

\frac{1}{2} k T

per molecule, or

\frac{1}{2} R T

per mole. So

U = \frac{f}{2} n R T

and

C_{v} = (1/ n) (d U / d T) = \frac{f}{2} R

Solved Example 1

During an expansion of an ideal gas, the work done by the gas is

100 J

and the heat capacity of the process is found to be

+ 2 J /^{\circ} C

. Find

Δ U

if the final temperature is

2 5^{\circ} C

higher than the initial.

Solution:

Heat absorbed: $q = C Δ T = 2 \times 25 = + 50 J$ .

Work done on gas (IUPAC): $w = - 100 J$ (since gas did $100 J$ of work).

First law: $Δ U = q + w = 50 + (- 100) = - 50 J$ .

5. The bomb calorimeter (constant volume)

Figure: Bomb calorimeter used to measure heat of combustion at constant volume. Heat released by combustion raises water temperature; measured

Δ T

combined with heat capacity of the system gives

Δ U

The bomb calorimeter, devised by Berthelot in 1881, measures the heat released in combustion at constant volume, giving $Δ U$ (not $Δ H$ ).

Construction

A thick-walled sealed steel container ("the bomb") that can withstand high pressure.
The bomb sits inside a well-insulated copper vessel filled with water.
A weighed sample is placed in a small dish inside the bomb along with oxygen at about $20 - 30 bar$ .
The water bath includes a stirrer (for uniform temperature) and a thermometer (accurate to $0.0 1^{\circ}$ C).
The whole apparatus is enclosed in an outer insulating jacket so no heat escapes.

Working

Record the initial water temperature.
Ignite the sample electrically. The combustion heat is transferred to the water and calorimeter.
Record the final water temperature. Calculate $Δ T$ .
Compute heat released: $q = C_{cal} Δ T$ , where $C_{cal}$ is the total heat capacity of the calorimeter (water + steel bomb + accessories), determined beforehand by burning a substance of known combustion heat.
Because $V$ is constant, this heat equals $Δ U$ of the sample (divided by moles for per-mole value).

Getting $Δ H$ from bomb calorimeter data

The bomb gives $Δ U$ ; use $Δ H = Δ U + Δ n_{g} R T$ to convert:

For combustion of a hydrocarbon:

Δ H = Δ U + Δ n_{g} R T

where

Δ n_{g}

is (moles of gaseous products) - (moles of gaseous reactants), ignoring solids and liquids.

Solved Example 2

A sample of

0.16 g

of CH

_{4}

is combusted at

2 7^{\circ} C

in a bomb calorimeter. The temperature of the system rises by

0. 5^{\circ} C

. Calculate the heat of combustion of methane at (i) constant volume and (ii) constant pressure. Thermal capacity of the calorimeter system is

17.7 kJ K^{- 1}

;

R = 8.314 J K^{- 1} mol^{- 1}

Solution:

(i) At constant volume:

Heat released by $0.16 g$ CH $_{4}$ : $q = C_{cal} Δ T = 17.7 \times 0.5 = 8.85 kJ$ .

Moles of CH $_{4}$ : $n = 0.16/16 = 0.01 mol$ .

Heat of combustion per mole: $Δ U = - \frac{8.85}{0.01} = - 885 kJ mol^{- 1}$ .

(ii) At constant pressure:

Balanced reaction: $CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (ℓ)$ .

$Δ n_{g} = 1 - 3 = - 2$ , $T = 300 K$ , $R = 8.314 \times 1 0^{- 3} kJ K^{- 1} mol^{- 1}$ .

$Δ H = Δ U + Δ n_{g} R T = - 885 + (- 2) (8.314 \times 1 0^{- 3}) (300)$

$= - 885 - 4.99 \approx - 889.99 kJ mol^{- 1}$ .

6. The coffee-cup calorimeter (constant pressure)

For reactions in solution (neutralization, dissolution, dilution) at atmospheric pressure, a much simpler apparatus is used - a coffee-cup calorimeter. It consists of a well-insulated polystyrene cup with a lid, a stirrer, and a thermometer.

Working

Measure the initial temperature $T_{1}$ of the solution in the cup.
Add the reactant (or mix the two solutions). Stir to complete the reaction quickly.
Record the maximum (or minimum) temperature $T_{2}$ . Compute $Δ T = T_{2} - T_{1}$ .
Heat absorbed by the solution: $q_{solution} = m s Δ T$ , using specific heat of the solution (approximately equal to that of water, $4.184 J g^{- 1} K^{- 1}$ , for dilute aqueous solutions).
Since $P$ is constant, the heat released by the reaction equals $Δ H$ : $Δ H = - q_{solution}$ .

Why constant pressure? The coffee cup is open to the atmosphere, so any small volume change of the solution is balanced by expansion into the air. External pressure remains at

P_{atm}

, so this measures

q_{p} = Δ H

directly - no

Δ n_{g} R T

correction needed.

Neutralization example

Mix $50 mL$ of $1.0 M HCl$ with $50 mL$ of $1.0 M NaOH$ (both at $25. 0^{\circ}$ C). Temperature rises to $31. 8^{\circ}$ C. Calculate $Δ H$ per mole of water formed.

Total mass $\approx 100 g$ ; $Δ T = 6. 8^{\circ}$ C; $s \approx 4.18 J g^{- 1} K^{- 1}$ .

Heat absorbed: $q = 100 \times 4.18 \times 6.8 = 2842 J$ .

Moles of water formed: $n = 0.050 \times 1.0 = 0.050 mol$ .

$Δ H = - \frac{2842}{0.050} = - 56800 J mol^{- 1} = - 56.8 kJ mol^{- 1}$ , close to the tabulated $- 57.1 kJ mol^{- 1}$ .

7. Water equivalent

The water equivalent of a calorimeter is the mass of water that would absorb the same heat as the calorimeter for the same temperature rise. If the calorimeter has heat capacity $C_{cal}$ :

Water equivalent:

W = \frac{C _{cal}}{s _{water}} = \frac{C _{cal}}{4.184 J g ^{- 1} K ^{- 1}}

(in grams)

This lets you treat the whole calorimeter as an equivalent extra mass of water, simplifying calculations when both water and calorimeter absorb heat.

8. Heat capacity for various processes

Since heat capacity depends on the process, different processes have different $C$ values for the same substance:

Process	Effective molar heat capacity
Isothermal ( $Δ T = 0$ , but $q \neq = 0$ )	$C = \pm \infty$
Isobaric	$C = C_{p}$
Isochoric	$C = C_{v}$
Adiabatic ( $q = 0$ )	$C = 0$

Depending on the process, $C$ can take any value from $- \infty$ to $+ \infty$ .

Solved Example 3

For Ag,

C_{p} (J K^{- 1} mol^{- 1}) = 24 + 0.006 T

. Calculate

Δ H

3 mol

of silver are raised from

2 7^{\circ} C

to its melting point

92 7^{\circ} C

under

1 atm

pressure.

Solution:

Since $C_{p}$ depends on $T$ , integrate: $Δ H = n \int_{T_{1}}^{T_{2}} C_{p} d T$ .

$T_{1} = 300 K$ , $T_{2} = 1200 K$ .

$\int_{300}^{1200} (24 + 0.006 T) d T = 24 (1200 - 300) + \frac{0.006}{2} (120 0^{2} - 30 0^{2})$

$= 24 \times 900 + 0.003 \times (1440000 - 90000) = 21600 + 4050 = 25650 J mol^{- 1}$ .

For $3 mol$ : $Δ H = 3 \times 25650 = 76950 J \approx 76.95 kJ$ .

9. Common Mistakes to Avoid

Watch out

Confusing $C_{p}$ with $C_{v}$ . Use $C_{p}$ for constant pressure processes ( $Δ H$ ); use $C_{v}$ for constant volume processes ( $Δ U$ ). Applying the wrong one for isobaric heating is a very common error.
Applying $C_{p} - C_{v} = R$ to non-ideal gases or condensed phases. This is Mayer's relation, valid only for ideal gases.
Forgetting that heat capacity is a path function. The heat capacity value depends on how the process is carried out (isothermal, adiabatic, isobaric, isochoric).
Assuming bomb calorimeter measures $Δ H$ . It measures $Δ U$ (constant volume). Use $Δ H = Δ U + Δ n_{g} R T$ to convert.
Not counting only gaseous species in $Δ n_{g}$ . Solids and liquids do not contribute to $Δ n_{g}$ since their molar volumes are negligible.
Confusing molar with specific heat. Molar: per mole (J mol $^{- 1}$ K $^{- 1}$ ). Specific: per gram (J g $^{- 1}$ K $^{- 1}$ ). Convert using $C = M s$ where $M$ is molar mass.
Ignoring the heat capacity of the calorimeter itself. In accurate work, $q_{total} = (m_{water} s_{water} + C_{cal}) Δ T$ - not just water alone.

Frequently Asked Questions

Q1. Why is $C_{p}$ greater than $C_{v}$ for a gas?

At constant pressure, adding heat raises the temperature and also does work by expanding the gas. So some of the heat goes into work, and you need more heat to achieve the same $Δ T$ . At constant volume, all the heat goes into internal energy. Hence $C_{p} > C_{v}$ , and the difference $C_{p} - C_{v} = R$ (Mayer's relation) is exactly the work per unit $T$ per mole.

Q2. Why does a bomb calorimeter measure $Δ U$ and not $Δ H$ ?

The bomb has a fixed steel volume - so the reaction proceeds at constant $V$ . By the first law with $w = 0$ (no volume change), the heat exchanged equals $Δ U$ . To obtain $Δ H$ , use $Δ H = Δ U + Δ n_{g} R T$ after the experiment.

Q3. Why does a coffee-cup calorimeter measure $Δ H$ directly?

The cup is open to the atmosphere, so the reaction occurs at constant atmospheric pressure. Under constant $P$ , the heat exchanged equals $Δ H$ . No conversion is needed - which is why coffee-cup calorimetry is the standard method for measuring enthalpies of neutralization, dissolution and dilution in solution.

Q4. What is the water equivalent of a calorimeter, and why is it used?

The water equivalent is the mass of water that would need the same amount of heat as the calorimeter to raise its temperature by the same amount. Adding it to the actual mass of water treats the whole system as one equivalent mass of water, simplifying $q = m s Δ T$ calculations.

Q5. Why is water used as the working fluid in calorimeters?

Water has a very high specific heat capacity ( $4.184 J g^{- 1} K^{- 1}$ ) - almost the highest of any common liquid. This means a small amount of water absorbs a lot of heat for a small temperature rise, which reduces heat losses to the surroundings and makes measurements more accurate.

Q6. Can heat capacity be negative or infinite? Give examples.

Yes - heat capacity is a path function and its value depends on the process. In an isothermal process $Δ T = 0$ while $q \neq = 0$ , so $C = q /Δ T$ becomes $\pm \infty$ . In an adiabatic process $q = 0$ while $Δ T \neq = 0$ , so $C = 0$ . Certain unusual processes can even give negative $C$ .

Q7. Why is $C_{v}$ for a monoatomic ideal gas exactly $\frac{3}{2} R$ ?

A monoatomic gas has only 3 translational degrees of freedom. By the equipartition theorem, each degree contributes $\frac{1}{2} k T$ per molecule, so total energy per molecule is $\frac{3}{2} k T$ , giving $U = \frac{3}{2} n R T$ . Then $C_{v} = (1/ n) (d U / d T) = \frac{3}{2} R$ . By Mayer's relation $C_{p} = \frac{5}{2} R$ .

Q8. In a bomb calorimeter experiment, why is oxygen loaded at high pressure?

To ensure complete combustion of the sample. If oxygen were limited, the sample could combust only partially - forming CO instead of CO $_{2}$ - giving a wrong (too low) heat of combustion value. Loading O $_{2}$ at $20 - 30 bar$ guarantees a large excess.

Previous year questions on Calorimetry

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

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Practice questions →Take a mock test

1. What is heat capacity

2. Three ways to express heat capacity

Total heat capacity (CT​)

Molar heat capacity (C)

Specific heat capacity (s)

Connecting the three

3. Cp​ and Cv​ - heat capacity depends on the process

Constant volume: heat = ΔU

Constant pressure: heat = ΔH

4. Mayer's relation: Cp​−Cv​=R

5. The bomb calorimeter (constant volume)

Construction

Working

Getting ΔH from bomb calorimeter data

6. The coffee-cup calorimeter (constant pressure)

Working

Neutralization example

7. Water equivalent

8. Heat capacity for various processes

9. Common Mistakes to Avoid

Frequently Asked Questions

Q1. Why is Cp​ greater than Cv​ for a gas?

Q2. Why does a bomb calorimeter measure ΔU and not ΔH?

Q3. Why does a coffee-cup calorimeter measure ΔH directly?

Q4. What is the water equivalent of a calorimeter, and why is it used?

Q5. Why is water used as the working fluid in calorimeters?

Q6. Can heat capacity be negative or infinite? Give examples.

Q7. Why is Cv​ for a monoatomic ideal gas exactly 23​R?

Q8. In a bomb calorimeter experiment, why is oxygen loaded at high pressure?

Previous year questions on Calorimetry

Ready to master Chemical Thermodynamics?

Total heat capacity $(C_{T})$

Molar heat capacity $(C)$

Specific heat capacity $(s)$

3. $C_{p}$ and $C_{v}$ - heat capacity depends on the process

Constant volume: heat = $Δ U$

Constant pressure: heat = $Δ H$

4. Mayer's relation: $C_{p} - C_{v} = R$

Getting $Δ H$ from bomb calorimeter data

Q1. Why is $C_{p}$ greater than $C_{v}$ for a gas?

Q2. Why does a bomb calorimeter measure $Δ U$ and not $Δ H$ ?

Q3. Why does a coffee-cup calorimeter measure $Δ H$ directly?

Q7. Why is $C_{v}$ for a monoatomic ideal gas exactly $\frac{3}{2} R$ ?