# How to determine the calorimeter constant from its heat capacity

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.

• I'd go with (A) - sum the heat capacity of the water (600 $\times$ 4.184) and the heat capacity of the calorimeter. – Todd Minehardt Oct 16 '15 at 15:13
• 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? – inspd Oct 17 '15 at 0:36
• See this Wikipedia article - "the magnitude of the degree Celsius is exactly equal to that of the kelvin." – Todd Minehardt Oct 17 '15 at 2:33

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.