A bomb calorimeter contains $600\;\mathrm{mL}$ of water. The calorimeter is calibrated electrically. The heat capacity of the calorimeter is $785\;\mathrm{J\,K^{-1}}$. The calorimeter constant would be closest to:

A. $3.29\;\mathrm{kJ\,ºC^{-1}}$

B. $4.18\;\mathrm{kJ\,ºC^{-1}}$

C. $4.97\;\mathrm{kJ\,ºC^{-1}}$

D. $789\;\mathrm{kJ\,ºC^{-1}}$

My (rather mindless) attempt is as follows: $$ E=mC_PT\to E/T=mC_P\to C_{\mathrm{cal}}=mC_P=(600)(8.314)(10^{-3})=4.9884\;\mathrm{kJ\,ºC^{-1}} $$ The closest answer to my result seems to be C ($4.97\;\mathrm{kJ\,ºC^{-1}}$), however I know I'm wrong.

  • $\begingroup$ I'd go with (A) - sum the heat capacity of the water (600 $\times$ 4.184) and the heat capacity of the calorimeter. $\endgroup$ Oct 16, 2015 at 15:13
  • $\begingroup$ But I don't understand how we can add $0.785 kj/K$ to $2.51 kj/ºC$ to obtain $3.29 kj/ºC$. Aren't they different units? $\endgroup$
    – inspd
    Oct 17, 2015 at 0:36
  • $\begingroup$ See this Wikipedia article - "the magnitude of the degree Celsius is exactly equal to that of the kelvin." $\endgroup$ Oct 17, 2015 at 2:33

1 Answer 1


To give a precise answer, the following assumptions are necessary and must be clear:

  1. bomb calorimeter works at constant volume ($V=const$);
  2. both water and calorimeter itself are at thermodynamic equilibrium before the experiment and during the measurement, in particular their temperatures $T_w$ and $T_c$ are equal before the experiment and during the measurement;
  3. the system is compound by calorimeter itself plus water;
  4. the system is an isolated one;
  5. pressure is 1 bar.

Initially the system is at temperature $T_1$. Let's imagine that an object at $T_o>T_1$ is put inside the chamber of the calorimeter. Temperature of the system increases and, once reached thermodynamic equilibrium, it stops at a precise value $T_2$.

Since $V=const$, heat transferred from object to system is: \begin{equation} Q_V=\Delta U=\Delta U_{calorimeter}+\Delta U_{water}=(mc_V\Delta T)_c+(mc_V\Delta T)_w \end{equation} where $\Delta T_c=\Delta T_w=T_2-T_1$.

We know that heat capacity at constant volume is defined as: \begin{equation} C_V=\left(\frac{\partial U}{\partial T}\right)_V\approx \left(\frac{\Delta U}{\Delta T}\right)_V \end{equation} So, reshaping the first equation, we obtain: \begin{equation} C_V=\frac{\Delta U}{\Delta T}=(mc_V)_c+(mc_V)_w=(C_V)_c+(\rho Vc_V)_w \end{equation} Adding the following data:

  1. $\rho_w=1000\;kg/m^3$;
  2. $(c_V(300\;K,1\;bar))_w\approx 4.134\;J/(kg\;K)$ (source: Perry's Chemical Engineers' Handbook)

and carrying out the conversion: $V=600\;mL=6\times10^{-4}\;m^3$, we obtain finally: \begin{equation} C_V=787\;J/K=0.787\;kJ/K \end{equation} So right answer is A.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.